Nonlinear Systems of Equations

48 soru

Soru 1Soru

In the xyxy-plane, the line y=x4y = x - 4 intersects the parabola y=x23x4y = x^2 - 3x - 4 at the points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). If x1<x2x_1 < x_2, what is the value of x2x_2?

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Cevap: 4

Cevap

4
The correct answer is 4. Setting the two equations equal to find their points of intersection gives x23x4=x4x^2 - 3x - 4 = x - 4. Subtracting xx and adding 44 to both sides yields the simplified quadratic equation x24x=0x^2 - 4x = 0. Factoring out xx gives x(x4)=0x(x - 4) = 0, which has the solutions x=0x = 0 and x=4x = 4. Given that x1<x2x_1 < x_2, we have x1=0x_1 = 0 and x2=4x_2 = 4. Therefore, the value of x2x_2 is 4.

Adım Adım Çözüm

1
Set the two equations equal to each other to find their points of intersection.
x23x4=x4x^2 - 3x - 4 = x - 4
At the intersection points, the y-values of the line and the parabola must be equal.
2
Subtract xx and add 44 to both sides of the equation to set it to zero.
x24x=0x^2 - 4x = 0
To solve a quadratic equation, we must rewrite it in standard form: ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic expression by factoring out the greatest common factor, which is xx.
x(x4)=0x(x - 4) = 0
Factoring allows us to use the zero product property to find the individual roots.
4
Solve for the two possible values of xx.
x=0x = 0 or x=4x = 4
Setting each factor to zero gives x=0x = 0 and x4=0x - 4 = 0, which simplifies to x=4x = 4.
5
Compare the two solutions to find the value of x2x_2 given the condition x1<x2x_1 < x_2.
x1=0x_1 = 0 and x2=4x_2 = 4
Since 0<40 < 4, the smaller value is x1x_1 and the larger value is x2x_2.

Anahtar Kavram

Solving a system of a linear equation and a quadratic equation by substitution.
Tahmini Süre:1m 0s
Soru 2Soru
y=x28x+cy=2x5\begin{aligned} y &= x^2 - 8x + c \\ y &= 2x - 5 \end{aligned}

In the system of equations above, cc is a constant. If the system has two real solutions, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), such that the product of the yy-coordinates of the solutions, y1y2y_1 y_2, is equal to 55, what is the value of cc?

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Cevap: 15

Cevap

The value of the constant cc is 1515.
Substituting the expression for yy from the linear equation into the quadratic equation yields the single variable quadratic equation x210x+(c+5)=0x^2 - 10x + (c+5) = 0. Using Vieta's formulas, the sum of the roots is x1+x2=10x_1 + x_2 = 10 and the product of the roots is x1x2=c+5x_1 x_2 = c+5. Substituting these relationships into the expanded product of the yy-coordinates, y1y2=(2x15)(2x25)=4x1x210(x1+x2)+25y_1 y_2 = (2x_1 - 5)(2x_2 - 5) = 4x_1 x_2 - 10(x_1 + x_2) + 25, allows us to set up the equation 5=4(c+5)100+255 = 4(c+5) - 100 + 25. Solving for cc yields 1515. Checking the discriminant of the quadratic equation at c=15c=15 gives 10080=20100 - 80 = 20, which is positive, confirming the existence of two distinct real solutions.

Adım Adım Çözüm

1
Substitute the expression for yy from the second equation into the first equation.
2x5=x28x+c    x210x+(c+5)=02x - 5 = x^2 - 8x + c \implies x^2 - 10x + (c+5) = 0
This substitution reduces the system to a single quadratic equation whose roots, x1x_1 and x2x_2, represent the xx-coordinates of the intersection points.
2
Apply Vieta's formulas to the resulting quadratic equation.
x1+x2=10x_1 + x_2 = 10 and x1x2=c+5x_1 x_2 = c+5
Vieta's formulas state that for a quadratic equation ax2+bx+d=0ax^2 + bx + d = 0, the sum of the roots is ba-\frac{b}{a} and the product of the roots is \frac{d}{a}.
3
Express the product of the yy-coordinates, y1y2y_1 y_2, in terms of x1x_1 and x2x_2 using the linear relationship.
y1y2=(2x15)(2x25)=4x1x210(x1+x2)+25y_1 y_2 = (2x_1 - 5)(2x_2 - 5) = 4x_1 x_2 - 10(x_1 + x_2) + 25
Since both intersection points lie on the line y=2x5y = 2x - 5, we can substitute y1=2x15y_1 = 2x_1 - 5 and y2=2x25y_2 = 2x_2 - 5 and expand.
4
Substitute the Vieta's formulas relations into the product equation and solve for cc.
5=4(c+5)10(10)+25    5=4c+20100+25    5=4c55    60=4c    c=155 = 4(c + 5) - 10(10) + 25 \implies 5 = 4c + 20 - 100 + 25 \implies 5 = 4c - 55 \implies 60 = 4c \implies c = 15
By substituting the known values of (x1+x2)(x_1 + x_2) and (x1x2)(x_1 x_2) and setting the product y1y2y_1 y_2 to 55, we obtain a linear equation in terms of cc that we can solve directly.

Anahtar Kavram

Solving systems of linear-quadratic equations using algebraic substitution and Vieta's formulas.
Soru 3Soru

If (x,y)(x, y) is the solution to the system of equations below, what is the value of xx?

xy=4x - y = 4
x2y2=40x^2 - y^2 = 40
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Cevap: 7

Cevap

The value of xx is 7.
The equation x2y2=40x^2 - y^2 = 40 can be factored as (xy)(x+y)=40(x-y)(x+y) = 40. Substituting xy=4x-y = 4 yields 4(x+y)=404(x+y) = 40, which simplifies to x+y=10x+y = 10. Adding the two linear equations xy=4x-y = 4 and x+y=10x+y = 10 eliminates yy, resulting in 2x=142x = 14, or x=7x = 7.

Adım Adım Çözüm

1
Factor the second equation using the difference of squares identity.
(xy)(x+y)=40(x-y)(x+y) = 40
To rewrite the quadratic expression in a form where the linear equation can be substituted.
2
Substitute the first equation xy=4x-y = 4 into the factored expression.
4(x+y)=404(x+y) = 40, which simplifies to x+y=10x+y = 10
To find a simpler linear relation for the sum of the variables.
3
Add the equations xy=4x-y = 4 and x+y=10x+y = 10.
2x=142x = 14
To eliminate the variable yy and solve for xx directly.
4
Solve for xx by dividing by 2.
x=7x = 7
To isolate the variable and find the final value of xx.

Anahtar Kavram

Solving a nonlinear system of equations by factoring a difference of squares and substituting.

Alternatif Yöntem

Express xx from the first equation as x=y+4x = y + 4 and substitute it into the second equation: (y+4)2y2=40(y+4)^2 - y^2 = 40. Expanding and simplifying yields y2+8y+16y2=40y^2 + 8y + 16 - y^2 = 40, which simplifies to 8y+16=40    8y=24    y=38y + 16 = 40 \implies 8y = 24 \implies y = 3. Substituting y=3y = 3 back into x=y+4x = y + 4 gives x=7x = 7.
Tahmini Süre:1m 15s
Soru 4Soru

A system of two equations is shown below.

y=x24y = x^2 - 4
y=x2y = x - 2

Which of the following ordered pairs (x,y)(x, y) is a solution to the system?

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Cevap: (2,0)(2, 0)

Cevap

The ordered pair (2,0)(2, 0) is the correct solution to the system.
The ordered pair (2,0)(2, 0) is the correct solution because substituting these values into both equations in the system results in true statements. For the quadratic equation, 0=2240 = 2^2 - 4 simplifies to 0=00 = 0. For the linear equation, 0=220 = 2 - 2 simplifies to 0=00 = 0.

Adım Adım Çözüm

1
Substitute the xx-value and yy-value from the candidate solution (2,0)(2, 0) into the first equation, y=x24y = x^2 - 4.
0=224    0=44    0=00 = 2^2 - 4 \implies 0 = 4 - 4 \implies 0 = 0, which is a true statement.
An ordered pair must satisfy the first equation to be a candidate solution for the system.
2
Substitute the xx-value and yy-value from the candidate solution (2,0)(2, 0) into the second equation, y=x2y = x - 2.
0=22    0=00 = 2 - 2 \implies 0 = 0, which is also a true statement.
An ordered pair must satisfy all equations in the system simultaneously to be a valid solution.

Anahtar Kavram

Verifying a solution to a nonlinear system of equations by substituting the coordinate values into both equations.
Soru 5Soru

If (x,y)(x, y) is a solution to the system of equations below and x>0x > 0, what is the value of xx?

y=x2y = x^2
y=x+6y = x + 6
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Cevap: 3

Cevap

The correct answer is 33.
Substituting y=x2y = x^2 into y=x+6y = x + 6 gives the quadratic equation x2x6=0x^2 - x - 6 = 0. Factoring this expression yields (x3)(x+2)=0(x - 3)(x + 2) = 0, which gives solutions of x=3x = 3 and x=2x = -2. Since the system requires x>0x > 0, the only valid solution is 33.

Adım Adım Çözüm

1
Substitute the expression for yy from the first equation into the second equation.
x2=x+6x^2 = x + 6
To eliminate the variable yy and solve for xx directly.
2
Subtract xx and 66 from both sides to write the quadratic equation in standard form.
x2x6=0x^2 - x - 6 = 0
Setting the quadratic expression equal to zero allows it to be factored.
3
Factor the quadratic trinomial.
(x3)(x+2)=0(x - 3)(x + 2) = 0
Finding factors whose product is 6-6 and whose sum is 1-1 helps find the roots.
4
Solve for xx and apply the constraint x>0x > 0.
x=3x = 3
The equation has solutions x=3x = 3 and x=2x = -2. Because xx must be greater than 00, we discard the negative solution.

Anahtar Kavram

Solving a system of nonlinear equations by substitution and factoring the resulting quadratic equation.
Soru 6Soru

The graphs of the equations y=x210y = x^2 - 10 and y=2x2y = 2x - 2 intersect at the point (x,y)(x, y) in the first quadrant. What is the value of yy?

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Cevap: 6

Cevap

The value of yy is 6.
Equating the equations gives x210=2x2x^2 - 10 = 2x - 2. Moving all terms to one side yields x22x8=0x^2 - 2x - 8 = 0, which factors as (x4)(x+2)=0(x - 4)(x + 2) = 0. Since the point is in the first quadrant, both coordinates must be positive, so we use x=4x = 4. Substituting x=4x = 4 into the linear equation gives y=2(4)2=6y = 2(4) - 2 = 6.

Adım Adım Çözüm

1
Equate the two expressions for yy
x210=2x2x^2 - 10 = 2x - 2
Since both equations define yy, their right-hand sides must be equal at the points of intersection.
2
Rewrite the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x22x8=0x^2 - 2x - 8 = 0
Subtract 2x2x and add 22 to both sides of the equation to set it equal to zero.
3
Factor the quadratic equation
(x4)(x+2)=0(x - 4)(x + 2) = 0
Find two integers that multiply to 8-8 and add to 2-2, which are 4-4 and 22.
4
Solve for the possible values of xx
x=4x = 4 or x=2x = -2
Set each factor equal to zero and solve.
5
Determine the positive xx-coordinate and find yy
y=6y = 6
For the point to be in the first quadrant, both coordinates must be positive. Thus, we select x=4x = 4 and substitute it into the linear equation: y=2(4)2=6y = 2(4) - 2 = 6.

Anahtar Kavram

Solving a system of linear and quadratic equations via substitution.

Alternatif Yöntem

Instead of factoring, the quadratic formula can be used to solve x22x8=0x^2 - 2x - 8 = 0: x=(2)±(2)24(1)(8)2(1)=2±362=2±62x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}. This yields x=4x = 4 and x=2x = -2. Then substitute the positive root to find yy.
Tahmini Süre:1m 30s
Soru 7Soru

A system of equations is shown below.

y=x22xy = x^2 - 2x
y=3y = 3

If (x,y)(x, y) is a solution to the system of equations and x>0x > 0, what is the value of x+yx + y?

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Cevap: 6

Cevap

6
To solve the system of equations, substitute the expression for yy from the second equation into the first equation: 3=x22x3 = x^2 - 2x. Subtracting 3 from both sides results in the quadratic equation x22x3=0x^2 - 2x - 3 = 0. Factoring the quadratic yields (x3)(x+1)=0(x - 3)(x + 1) = 0, which gives the possible values of xx as 33 and 1-1. Since the problem specifies that x>0x > 0, the value of xx is 3. Given that y=3y = 3, the value of x+yx + y is 3+3=63 + 3 = 6.

Adım Adım Çözüm

1
Substitute the value of yy from the second equation into the first equation.
3=x22x3 = x^2 - 2x
Since both equations are equal to yy, their right-hand sides must be equal to each other.
2
Rearrange the equation to set it equal to zero.
x22x3=0x^2 - 2x - 3 = 0
This puts the equation into standard quadratic form so that it can be factored.
3
Factor the quadratic equation.
(x3)(x+1)=0(x - 3)(x + 1) = 0
Finding factors of -3 that add up to -2 helps isolate the solutions for xx.
4
Find the solutions for xx and apply the given constraint.
x=3x = 3 (since x>0x > 0)
The factors give x=3x = 3 and x=1x = -1. The constraint x>0x > 0 excludes x=1x = -1.
5
Calculate the value of x+yx + y.
3+3=63 + 3 = 6
Substitute the value of x=3x = 3 and the given value of y=3y = 3 to find the final sum.

Anahtar Kavram

Solving nonlinear systems of equations using substitution and solving quadratic equations by factoring.
Soru 8Soru

In the system of equations below, kk is a constant.

y=3x25x+4y = 3x^2 - 5x + 4
y=7x+ky = 7x + k

If the system has exactly one real solution (x,y)(x, y), what is the value of kk?

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Cevap: 8-8

Cevap

8-8
To find the number of solutions for the system of equations, we set the equations equal to each other: 3x25x+4=7x+k3x^2 - 5x + 4 = 7x + k. Subtracting 7x7x and kk from both sides gives the standard quadratic equation 3x212x+(4k)=03x^2 - 12x + (4 - k) = 0. For a quadratic equation to have exactly one real solution, its discriminant, b24acb^2 - 4ac, must equal 00. Substituting a=3a = 3, b=12b = -12, and c=4kc = 4 - k into the discriminant formula gives (12)24(3)(4k)=0(-12)^2 - 4(3)(4 - k) = 0. Simplifying this equation yields 14412(4k)=0144 - 12(4 - k) = 0, which simplifies to 14448+12k=0144 - 48 + 12k = 0, or 96+12k=096 + 12k = 0. Solving for kk gives k=8k = -8.

Adım Adım Çözüm

1
Equate the two equations to eliminate yy and form a single quadratic equation.
3x25x+4=7x+k3x212x+(4k)=03x^2 - 5x + 4 = 7x + k \Rightarrow 3x^2 - 12x + (4 - k) = 0
To find the points of intersection of the system, we set the expressions for yy equal to each other.
2
Identify the coefficients of the quadratic equation in standard form ax2+bx+c=0ax^2 + bx + c = 0.
a=3a = 3, b=12b = -12, and c=4kc = 4 - k
These coefficients are needed to calculate the discriminant of the quadratic equation.
3
Set the discriminant Δ=b24ac\Delta = b^2 - 4ac equal to zero and solve for kk.
(12)24(3)(4k)=014412(4k)=012(4k)=1444k=12k=8(-12)^2 - 4(3)(4 - k) = 0 \Rightarrow 144 - 12(4 - k) = 0 \Rightarrow 12(4 - k) = 144 \Rightarrow 4 - k = 12 \Rightarrow k = -8
A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

Anahtar Kavram

Solving nonlinear systems of equations by setting them equal to each other and using the discriminant of the resulting quadratic equation to determine the number of solutions.
Tahmini Süre:1m 30s
Soru 9Soru

A linear equation and a quadratic equation form a system, as shown.

y=x2+5y = x^2 + 5
y=3x+5y = 3x + 5

If the ordered pair (x,y)(x, y) is a solution to the system where xx is positive, what is the value of xx?

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Cevap: 3

Cevap

The value of xx is 3.
Equating the two expressions for yy gives the equation x2+5=3x+5x^2 + 5 = 3x + 5. Subtracting 5 from both sides results in x2=3xx^2 = 3x. Subtracting 3x3x from both sides gives the quadratic equation x23x=0x^2 - 3x = 0, which can be factored as x(x3)=0x(x - 3) = 0. This yields two solutions for xx: 00 and 33. Since the question specifies that xx is positive, the correct value is 3.

Adım Adım Çözüm

1
Equate the equations
x2+5=3x+5x^2 + 5 = 3x + 5
Since both equations are solved for yy, we can substitute the quadratic expression into the linear equation.
2
Simplify the equation
x23x=0x^2 - 3x = 0
Subtracting 5 from both sides and then subtracting 3x3x from both sides collects all terms on one side.
3
Factor and solve
x(x3)=0x(x - 3) = 0, so x=0x = 0 or x=3x = 3
Factoring out the greatest common factor, xx, allows us to apply the zero product property to find the individual roots.
4
Apply the constraint
x=3x = 3
The problem states that xx must be positive, which excludes the solution x=0x = 0.

Anahtar Kavram

Solving a system consisting of a linear equation and a quadratic equation by substitution.
Soru 10Soru

In the xyxy-plane, the line y=mxy = mx, where mm is a positive constant, is tangent to the parabola y=x2+9y = x^2 + 9. What is the value of mm?

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Cevap: 6

Cevap

The value of mm is 6.
Equating the equations of the line and the parabola yields the quadratic equation x2mx+9=0x^2 - mx + 9 = 0. For the line to be tangent to the parabola, this system must have exactly one real solution, meaning the discriminant b24acb^2 - 4ac must equal 00. Substituting a=1a = 1, b=mb = -m, and c=9c = 9 into the discriminant formula gives (m)24(1)(9)=0(-m)^2 - 4(1)(9) = 0, which simplifies to m236=0m^2 - 36 = 0. Solving for the positive constant mm gives 66.

Adım Adım Çözüm

1
Set the equation of the line equal to the equation of the parabola to find their intersection points.
mx=x2+9mx = x^2 + 9
The intersection points of the system correspond to the values of xx where both equations have the same yy-value.
2
Rewrite the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0 by subtracting mxmx from both sides.
x2mx+9=0x^2 - mx + 9 = 0
This allows the identification of the coefficients aa, bb, and cc to compute the discriminant.
3
Identify the coefficients and set the discriminant b24acb^2 - 4ac equal to 0.
(m)24(1)(9)=0(-m)^2 - 4(1)(9) = 0
A line is tangent to a parabola if and only if the system has exactly one real solution, which corresponds to a discriminant of zero.
4
Solve the equation for the positive constant mm.
m236=0    m2=36    m=6m^2 - 36 = 0 \implies m^2 = 36 \implies m = 6
Taking the square root of both sides gives m=±6m = \pm 6. Since mm is specified as a positive constant, we choose m=6m = 6.

Anahtar Kavram

Solving nonlinear systems of equations where a line is tangent to a parabola by setting the discriminant of the resulting quadratic equation to zero.
Soru 11Soru

A circle and a line are graphed in the xyxy-plane. The circle is defined by the equation x2+y2=25x^2 + y^2 = 25, and the line is defined by the equation y=3y = 3. If the line intersects the circle at the point (x,3)(x, 3), where x>0x > 0, what is the value of xx?

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Cevap: 4

Cevap

The correct answer is 4.
Substituting y=3y = 3 into the circle equation x2+y2=25x^2 + y^2 = 25 yields x2+32=25x^2 + 3^2 = 25. Simplifying this gives x2+9=25x^2 + 9 = 25, which simplifies to x2=16x^2 = 16. Taking the square root of both sides gives x=4x = 4 or x=4x = -4. Since it is given that x>0x > 0, the value of xx must be 44.

Adım Adım Çözüm

1
Substitute the value of y=3y = 3 into the circle's equation.
x2+32=25x^2 + 3^2 = 25
Since the line is y=3y = 3, any point of intersection must satisfy this y-coordinate. Substituting it into the circle's equation allows us to solve for the x-coordinate.
2
Simplify the equation and isolate x2x^2.
x2=16x^2 = 16
Squaring 33 gives 99, and subtracting 99 from both sides of the equation x2+9=25x^2 + 9 = 25 isolates x2x^2.
3
Solve for xx and apply the constraint x>0x > 0.
x=4x = 4
Taking the square root of both sides of x2=16x^2 = 16 gives x=4x = 4 or x=4x = -4. Since the problem states x>0x > 0, the only valid solution is 44.

Anahtar Kavram

Solving systems of nonlinear equations using substitution
Soru 12Soru

A system of equations consists of the equations y=x2+2x+7y = -x^2 + 2x + 7 and y=6x+ky = 6x + k, where kk is a constant. If the system has two distinct real solutions, what is the greatest integer value of kk?

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Cevap: 10

Cevap

The correct answer is 10. The greatest integer value of the constant that allows the system to have two distinct real solutions is 10.
To find the number of solutions to the system, equate the two equations: x2+2x+7=6x+k-x^2 + 2x + 7 = 6x + k. Rearranging this equation into standard quadratic form gives x2+4x+(k7)=0x^2 + 4x + (k - 7) = 0. For the system to have two distinct real solutions, the discriminant of this quadratic equation must be strictly greater than zero. The discriminant is calculated as b24ac=424(1)(k7)=164k+28=444kb^2 - 4ac = 4^2 - 4(1)(k - 7) = 16 - 4k + 28 = 44 - 4k. Setting this greater than zero yields 444k>044 - 4k > 0, which simplifies to k<11k < 11. The greatest integer value of kk that is strictly less than 11 is 10.

Adım Adım Çözüm

1
Equate the expressions for yy from both equations.
x2+2x+7=6x+k-x^2 + 2x + 7 = 6x + k
At the points of intersection, the yy-values of both equations must be equal.
2
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2+4x+(k7)=0x^2 + 4x + (k - 7) = 0
Standard form is required to calculate the discriminant of the quadratic equation.
3
Write the expression for the discriminant Δ=b24ac\Delta = b^2 - 4ac using the coefficients from the quadratic equation.
Δ=424(1)(k7)=444k\Delta = 4^2 - 4(1)(k - 7) = 44 - 4k
The discriminant determines the number of real solutions to the quadratic equation.
4
Set the discriminant to be strictly greater than 0 and solve the inequality for kk.
444k>0    k<1144 - 4k > 0 \implies k < 11
For the system to have two distinct real solutions, the discriminant must be positive.
5
Determine the greatest integer value of kk that satisfies the inequality k<11k < 11.
10
The largest integer strictly less than 11 is 10.

Anahtar Kavram

Using the discriminant of a quadratic equation derived from a nonlinear system to determine the number of real solutions.
Soru 13Soru
y=x27y=2x+1\begin{aligned} y &= x^2 - 7 \\ y &= 2x + 1 \end{aligned}

If (x,y)(x, y) is a solution to the system of equations above and y>0y > 0, what is the value of yy?

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Cevap: 9

Cevap

9
Substituting the expression for yy from the second equation into the first equation yields 2x+1=x272x + 1 = x^2 - 7. Subtracting 2x2x and 11 from both sides results in the standard form quadratic equation x22x8=0x^2 - 2x - 8 = 0. Factoring this equation gives (x4)(x+2)=0(x - 4)(x + 2) = 0, which yields the solutions x=4x = 4 and x=2x = -2. Substituting these xx-values back into the linear equation y=2x+1y = 2x + 1 gives the corresponding yy-values: y=2(4)+1=9y = 2(4) + 1 = 9 and y=2(2)+1=3y = 2(-2) + 1 = -3. Since the system specifies the constraint y>0y > 0, the correct value of yy must be 99.

Adım Adım Çözüm

1
Substitute the expression for yy from the linear equation into the quadratic equation.
2x+1=x272x + 1 = x^2 - 7
This eliminates the variable yy and leaves a single quadratic equation in terms of xx.
2
Rearrange the quadratic equation into standard form ax2+bx+c=0ax^2 + bx + c = 0.
x22x8=0x^2 - 2x - 8 = 0
Moving all terms to one side allows us to solve the quadratic equation by factoring.
3
Factor the quadratic equation.
(x4)(x+2)=0(x - 4)(x + 2) = 0, which gives x=4x = 4 or x=2x = -2.
Finding the roots of the quadratic equation provides the possible xx-coordinates of the solution points.
4
Substitute the xx-values back into the linear equation y=2x+1y = 2x + 1 to find the corresponding yy-values.
For x=4x = 4, y=2(4)+1=9y = 2(4) + 1 = 9. For x=2x = -2, y=2(2)+1=3y = 2(-2) + 1 = -3.
This gives the complete coordinate pairs (4,9)(4, 9) and (2,3)(-2, -3) for the system's solutions.
5
Apply the constraint y>0y > 0 to identify the correct solution.
Since 9>09 > 0 and 3<0-3 < 0, the correct value of yy is 99.
Only the solution (4,9)(4, 9) satisfies the condition that yy must be positive.

Anahtar Kavram

Solving linear-quadratic systems of equations by substitution and evaluating solutions under given constraints.
Soru 14Soru

A circle in the xyxy-plane is defined by the equation (x2)2+(y+1)2=10(x - 2)^2 + (y + 1)^2 = 10. The line y=3x+ky = 3x + k, where kk is a constant, is tangent to the circle. If k<0k < 0, what is the value of kk?

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Cevap: -17

Cevap

The correct value of kk is 17-17.
Substituting the line equation y=3x+ky = 3x + k into the circle equation yields (x2)2+(3x+k+1)2=10(x - 2)^2 + (3x + k + 1)^2 = 10. Expanding and writing this in standard form gives 10x2+(6k+2)x+(k2+2k5)=010x^2 + (6k + 2)x + (k^2 + 2k - 5) = 0. For the line to be tangent to the circle, the quadratic equation must have exactly one real solution, meaning its discriminant must be 00. Setting the discriminant Δ=(6k+2)24(10)(k2+2k5)\Delta = (6k+2)^2 - 4(10)(k^2 + 2k - 5) to 00 and simplifying gives 4k256k+204=0-4k^2 - 56k + 204 = 0. Dividing by 4-4 yields k2+14k51=0k^2 + 14k - 51 = 0, which factors as (k+17)(k3)=0(k+17)(k-3)=0. Since k<0k < 0, the value of kk must be 17-17.

Adım Adım Çözüm

1
Substitute the linear equation y=3x+ky = 3x + k into the circle equation.
(x2)2+(3x+k+1)2=10(x - 2)^2 + (3x + k + 1)^2 = 10
To find the coordinates where the line and the circle intersect.
2
Expand both squared terms and simplify the equation to standard quadratic form Ax2+Bx+C=0Ax^2 + Bx + C = 0.
10x2+(6k+2)x+(k2+2k5)=010x^2 + (6k + 2)x + (k^2 + 2k - 5) = 0
To write the system as a single quadratic equation in terms of xx.
3
Set the discriminant of the quadratic equation to zero.
(6k+2)24(10)(k2+2k5)=0(6k + 2)^2 - 4(10)(k^2 + 2k - 5) = 0 which simplifies to 4k256k+204=0-4k^2 - 56k + 204 = 0
Since the line is tangent to the circle, there must be exactly one intersection point, which means the quadratic equation must have exactly one real solution.
4
Divide the simplified equation by 4-4 and solve for kk.
k2+14k51=0    (k+17)(k3)=0    k=17 or k=3k^2 + 14k - 51 = 0 \implies (k + 17)(k - 3) = 0 \implies k = -17 \text{ or } k = 3
To find the values of kk that make the line tangent to the circle.
5
Apply the given constraint k<0k < 0.
k=17k = -17
The problem specifies that kk must be a negative value.

Anahtar Kavram

Solving nonlinear systems of equations involving circles and lines by substitution and using the discriminant to determine tangency.
Tahmini Süre:2m 30s
Soru 15Soru

A circle in the xyxy-plane is defined by the equation (xh)2+(yk)2=16(x - h)^2 + (y - k)^2 = 16, where hh and kk are constants. The center (h,k)(h, k) of the circle lies on the line y=xy = x. A second line, which passes through the origin and has a slope of 34-\frac{3}{4}, is tangent to the circle at exactly one point (x,y)(x, y). If this point of tangency lies in a quadrant where x>0x > 0 and y<0y < 0, what is the value of hh?

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Cevap: 207\frac{20}{7}

Cevap

207\frac{20}{7}
The correct option is 207\frac{20}{7}. By setting the center of the circle to (h,h)(h, h) and the equation of the line to 3x+4y=03x + 4y = 0, we find that the distance from the center to the line is 7h5\frac{|7h|}{5}. Since the line is tangent to the circle, this distance must equal the radius, which is 44. This gives h=±207h = \pm\frac{20}{7}. Finding the point of tangency shows that x=425hx = \frac{4}{25}h and y=325hy = -\frac{3}{25}h. For the point of tangency to lie in Quadrant IV (x>0x > 0 and y<0y < 0), hh must be positive, which yields h=207h = \frac{20}{7}.

Adım Adım Çözüm

1
Express the center of the circle and the equation of the tangent line in terms of the given parameters.
Since the center (h,k)(h, k) lies on the line y=xy = x, we have k=hk = h. The circle has radius R=16=4R = \sqrt{16} = 4 and is centered at (h,h)(h, h). The tangent line passes through the origin with slope 34-\frac{3}{4}, so its equation is y=34xy = -\frac{3}{4}x, which simplifies to 3x+4y=03x + 4y = 0.
Setting up the algebraic expressions for both geometric entities is necessary to relate them using coordinate geometry formulas.
2
Apply the tangency condition using the point-to-line distance formula.
The distance from the center (h,h)(h, h) to the line 3x+4y=03x + 4y = 0 must equal the radius 44. Thus: 3h+4h32+42=4    7h5=4    7h=20\frac{|3h + 4h|}{\sqrt{3^2 + 4^2}} = 4 \implies \frac{|7h|}{5} = 4 \implies |7h| = 20.
A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line equals the radius.
3
Solve for the possible values of hh.
h=±207h = \pm\frac{20}{7}.
Solving the absolute value equation yields two symmetric possibilities for the x-coordinate of the circle's center.
4
Determine the relationship between the center hh and the coordinates of the point of tangency (x,y)(x, y) to apply the quadrant constraint.
The radius connecting the center (h,h)(h, h) to the point of tangency (x,y)(x, y) is perpendicular to the tangent line. Since the tangent line has a slope of 34-\frac{3}{4}, the perpendicular radius line has a slope of 43\frac{4}{3}. Its equation is: yh=43(xh)    y=43x13hy - h = \frac{4}{3}(x - h) \implies y = \frac{4}{3}x - \frac{1}{3}h.
The intersection of the perpendicular radius line and the tangent line will locate the exact point of tangency.
5
Solve the system of equations for the point of tangency (x,y)(x, y) in terms of hh.
Equating the tangent line and the perpendicular line: 34x=43x13h    912x=1612x412h    2512x=412h    x=425h-\frac{3}{4}x = \frac{4}{3}x - \frac{1}{3}h \implies -\frac{9}{12}x = \frac{16}{12}x - \frac{4}{12}h \implies -\frac{25}{12}x = -\frac{4}{12}h \implies x = \frac{4}{25}h. Substituting back: y=34(425h)=325hy = -\frac{3}{4}\left(\frac{4}{25}h\right) = -\frac{3}{25}h.
This yields the coordinates of the tangency point as a function of the parameter hh.
6
Apply the quadrant constraint (x>0x > 0 and y<0y < 0) to choose the correct sign of hh.
We require x=425h>0x = \frac{4}{25}h > 0 and y=325h<0y = -\frac{3}{25}h < 0. Both inequalities are satisfied if and only if h>0h > 0. Therefore, h=207h = \frac{20}{7}.
This filters out the extraneous geometric solution that lies in Quadrant II.

Anahtar Kavram

Solving systems of nonlinear equations representing circles and lines by utilizing geometric relations, distance formulas, and quadrant constraints.
Tahmini Süre:3m 0s
Soru 16Soru

The graphs of the equations y2x=3y - 2x = 3 and y=x2y = x^2 intersect at two points in the xyxy-plane. If (x,y)(x, y) represents an intersection point with a positive xx-coordinate, what is the value of yy?

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Cevap: 9

Cevap

9
Substituting y=x2y = x^2 into the first equation yields x22x=3x^2 - 2x = 3. Setting the quadratic equation to zero gives x22x3=0x^2 - 2x - 3 = 0, which factors as (x3)(x+1)=0(x - 3)(x + 1) = 0. This gives the solutions x=3x = 3 and x=1x = -1. Because the question specifies a positive xx-coordinate, xx must be 3. Substituting x=3x = 3 back into y=x2y = x^2 yields y=9y = 9. Thus, the correct answer is 9.

Adım Adım Çözüm

1
Substitute y=x2y = x^2 into the equation y2x=3y - 2x = 3.
x22x=3x^2 - 2x = 3
This eliminates the variable yy and creates a single quadratic equation in terms of xx to find the xx-coordinates of the intersection points.
2
Rewrite the quadratic equation in standard form.
x22x3=0x^2 - 2x - 3 = 0
Subtracting 3 from both sides sets the quadratic equation to 0, which is necessary for factoring.
3
Factor the quadratic equation to find its solutions.
(x3)(x+1)=0(x - 3)(x + 1) = 0, which gives x=3x = 3 or x=1x = -1.
Factoring allows us to find the roots, which represent the xx-coordinates of the intersection points.
4
Apply the constraint that the xx-coordinate must be positive.
x=3x = 3
The question specifies that x>0x > 0, so we discard x=1x = -1.
5
Substitute x=3x = 3 back into y=x2y = x^2 to find the corresponding value of yy.
y=32=9y = 3^2 = 9
This determines the yy-coordinate of the intersection point with the positive xx-coordinate.

Anahtar Kavram

Solving a system of a linear equation and a quadratic equation by substitution.
Soru 17Soru

In the system of equations below, kk is a positive constant.

xy=kx - y = k
x23xy+y2=5x^2 - 3xy + y^2 = 5

If the system has exactly one real solution (x,y)(x, y), what is the value of kk?

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Cevap: 2

Cevap

The value of kk is 2.
Substituting y=xky = x - k into the second equation yields x2+kx+k25=0-x^2 + kx + k^2 - 5 = 0, which can be rewritten in standard form as x2kx+(5k2)=0x^2 - kx + (5 - k^2) = 0. For this quadratic equation to have exactly one real solution, its discriminant must equal 0: b24ac=(k)24(1)(5k2)=5k220=0b^2 - 4ac = (-k)^2 - 4(1)(5 - k^2) = 5k^2 - 20 = 0. Solving 5k220=05k^2 - 20 = 0 gives k2=4k^2 = 4, and since kk must be positive, k=2k = 2.

Adım Adım Çözüm

1
Rearrange the first equation to express yy in terms of xx.
y=xky = x - k
This allows for substitution into the second equation to eliminate yy.
2
Substitute y=xky = x - k into the second equation and expand.
x23x(xk)+(xk)2=5    x2+kx+k25=0x^2 - 3x(x - k) + (x - k)^2 = 5 \implies -x^2 + kx + k^2 - 5 = 0
To create a single quadratic equation in terms of xx.
3
Multiply by 1-1 to write the quadratic equation in standard form ax2+bx+c=0ax^2 + bx + c = 0.
x2kx+(5k2)=0x^2 - kx + (5 - k^2) = 0
Standard form makes it easier to identify the coefficients a=1a = 1, b=kb = -k, and c=5k2c = 5 - k^2.
4
Set the discriminant b24acb^2 - 4ac equal to 0.
(k)24(1)(5k2)=0(-k)^2 - 4(1)(5 - k^2) = 0
A quadratic equation has exactly one real solution if and only if its discriminant is zero.
5
Simplify the discriminant equation and solve for kk.
5k220=0    k2=4    k=25k^2 - 20 = 0 \implies k^2 = 4 \implies k = 2 (since kk must be positive)
To find the positive constant kk that satisfies the condition.

Anahtar Kavram

Determining the number of solutions to a nonlinear system by substituting and setting the discriminant of the resulting quadratic equation to zero.

Alternatif Yöntem

Alternatively, one can rewrite the second equation by grouping: x23xy+y2=(xy)2xy=5x^2 - 3xy + y^2 = (x - y)^2 - xy = 5. Since xy=kx - y = k, we have k2xy=5k^2 - xy = 5, so xy=k25xy = k^2 - 5. We now have a system of xy=kx - y = k and xy=k25xy = k^2 - 5. Substituting y=xky = x - k gives x(xk)=k25x(x - k) = k^2 - 5, leading to x2kx+(5k2)=0x^2 - kx + (5 - k^2) = 0, which can be solved using the discriminant as shown in the primary method.
Tahmini Süre:2m 30s
Soru 18Soru

A parabola and a line intersect at exactly one point in the xyxy-plane. The parabola is defined by the equation y=x2+6x2y = -x^2 + 6x - 2 and the line is defined by the equation y=2x+ky = 2x + k, where kk is a constant. What is the value of kk?

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Cevap: 2

Cevap

The correct answer is 2.
To find the value of kk where the parabola and the line intersect at exactly one point, we equate the two equations: x2+6x2=2x+k-x^2 + 6x - 2 = 2x + k. Rearranging this into standard quadratic form yields x24x+(k+2)=0x^2 - 4x + (k + 2) = 0. For a quadratic equation to have exactly one real solution, its discriminant, b24acb^2 - 4ac, must be equal to zero. Substituting a=1a = 1, b=4b = -4, and c=k+2c = k + 2 into the discriminant formula gives (4)24(1)(k+2)=0(-4)^2 - 4(1)(k + 2) = 0, which simplifies to 164k8=016 - 4k - 8 = 0, or 84k=08 - 4k = 0. Solving for kk gives k=2k = 2.

Adım Adım Çözüm

1
Equate the equations of the parabola and the line to set up an equation for the x-coordinates of their intersection points.
x2+6x2=2x+k-x^2 + 6x - 2 = 2x + k
At the points of intersection, the y-values of both equations are equal.
2
Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x24x+(k+2)=0x^2 - 4x + (k + 2) = 0
This allows us to identify the coefficients a=1a = 1, b=4b = -4, and c=k+2c = k + 2 to apply the quadratic discriminant.
3
Set the discriminant b24acb^2 - 4ac equal to zero.
(4)24(1)(k+2)=0(-4)^2 - 4(1)(k + 2) = 0
A quadratic system has exactly one real solution (tangency) if and only if the discriminant of the resulting quadratic equation is zero.
4
Solve the linear equation for kk.
k=2k = 2
Simplifying the expression yields 164k8=016 - 4k - 8 = 0, which simplifies to 84k=08 - 4k = 0, giving k=2k = 2.

Anahtar Kavram

Nonlinear Systems of Equations
Soru 19Soru
y=x25x+8y = x^2 - 5x + 8
y=2x+2y = 2x + 2

The system of equations above has two real solutions, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). If y1>y2y_1 > y_2, what is the value of x1x2x_1 - x_2?

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Cevap: 5

Cevap

The value of the difference between the two x-coordinates is 5.
To solve the system, substitute y=2x+2y = 2x + 2 into y=x25x+8y = x^2 - 5x + 8 to obtain 2x+2=x25x+82x + 2 = x^2 - 5x + 8. Subtracting 2x+22x + 2 from both sides gives the quadratic equation x27x+6=0x^2 - 7x + 6 = 0. Factoring this equation yields (x6)(x1)=0(x - 6)(x - 1) = 0, which gives the x-coordinates x=6x = 6 and x=1x = 1. Substituting these back into y=2x+2y = 2x + 2 gives the corresponding y-coordinates: y=14y = 14 when x=6x = 6, and y=4y = 4 when x=1x = 1. Thus, the two solutions are (6,14)(6, 14) and (1,4)(1, 4). Because y1>y2y_1 > y_2, we must have (x1,y1)=(6,14)(x_1, y_1) = (6, 14) and (x2,y2)=(1,4)(x_2, y_2) = (1, 4). The value of x1x2x_1 - x_2 is therefore 61=56 - 1 = 5.

Adım Adım Çözüm

1
Substitute the expression for yy from the second equation into the first equation.
2x+2=x25x+82x + 2 = x^2 - 5x + 8
This substitution reduces the system of two equations to a single quadratic equation in terms of xx.
2
Rearrange the quadratic equation into standard form by subtracting 2x2x and 22 from both sides.
x27x+6=0x^2 - 7x + 6 = 0
Putting the equation in standard form is necessary to factor it and find its roots.
3
Factor the quadratic equation to find the two possible values of xx.
(x6)(x1)=0(x - 6)(x - 1) = 0, which gives x=6x = 6 or x=1x = 1.
Factoring allows us to find the x-coordinates of the points where the two graphs intersect.
4
Substitute the x-values back into the linear equation y=2x+2y = 2x + 2 to find their corresponding y-values.
For x=6x = 6, y=2(6)+2=14y = 2(6) + 2 = 14. For x=1x = 1, y=2(1)+2=4y = 2(1) + 2 = 4. The two solutions are (6,14)(6, 14) and (1,4)(1, 4).
Finding the y-values helps identify which coordinate pair corresponds to (x1,y1)(x_1, y_1) and which to (x2,y2)(x_2, y_2) using the given condition.
5
Apply the condition y1>y2y_1 > y_2 to assign the variables and calculate x1x2x_1 - x_2.
Since 14>414 > 4, the solution with the larger y-value is (x1,y1)=(6,14)(x_1, y_1) = (6, 14) and the other is (x2,y2)=(1,4)(x_2, y_2) = (1, 4). Thus, x1x2=61=5x_1 - x_2 = 6 - 1 = 5.
This calculation yields the final requested value.

Anahtar Kavram

Solving a system of a linear equation and a quadratic equation by substitution.
Soru 20Soru
y=x25y=4x\begin{aligned} y &= x^2 - 5 \\ y &= 4x \end{aligned}

If (x,y)(x, y) is a solution to the system of equations above and y>0y > 0, what is the value of yy?

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Cevap: 20

Cevap

20
Substituting y=4xy = 4x into the equation y=x25y = x^2 - 5 yields 4x=x254x = x^2 - 5. Rearranging this equation into standard form gives x24x5=0x^2 - 4x - 5 = 0. Factoring the quadratic expression gives (x5)(x+1)=0(x - 5)(x + 1) = 0, which means x=5x = 5 or x=1x = -1. Substituting these values back into y=4xy = 4x gives the coordinates of the two solutions: (5,20)(5, 20) and (1,4)(-1, -4). Since the question specifies that y>0y > 0, the correct value is the positive yy-value, which is 20.

Adım Adım Çözüm

1
Substitute the expression for yy from the second equation into the first equation.
4x=x254x = x^2 - 5
This eliminates yy and creates a single equation in terms of xx.
2
Subtract 4x4x from both sides to write the quadratic equation in standard form.
x24x5=0x^2 - 4x - 5 = 0
Standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0 is required to solve by factoring.
3
Factor the quadratic equation.
(x5)(x+1)=0(x - 5)(x + 1) = 0
Factoring helps find the values of xx that satisfy the equation.
4
Solve for xx by setting each factor to zero.
x=5x = 5 or x=1x = -1
Applying the zero product property yields the possible xx-coordinates of the solutions.
5
Calculate the corresponding yy-values using the equation y=4xy = 4x.
For x=5x = 5, y=20y = 20. For x=1x = -1, y=4y = -4.
This determines the coordinates of the intersection points, which are (5,20)(5, 20) and (1,4)(-1, -4).
6
Apply the constraint y>0y > 0 to identify the correct value of yy.
y=20y = 20
Since 4-4 is not greater than zero, the only valid solution is (5,20)(5, 20), giving y=20y = 20.

Anahtar Kavram

Solving a system of linear and quadratic equations using substitution.
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Nonlinear Systems of Equations Alıştırma Soruları — SAT | Examkin