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Simultaneous Linear and Quadratic Equations

26 questions

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Question 21Question →

Find the positive value of xx that satisfies the simultaneous equations y−2x=3y - 2x = 3 and y=x2+3y = x^2 + 3.

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Answer: 22

Answer

The positive value of xx is 2.
Rearranging the linear equation gives y=2x+3y = 2x + 3. Setting this equal to the second expression for yy gives 2x+3=x2+32x + 3 = x^2 + 3, which simplifies to x2−2x=0x^2 - 2x = 0. Factoring out xx yields x(x−2)=0x(x - 2) = 0. The solutions for xx are 00 and 22. The positive solution is 22.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation
y=2x+3y = 2x + 3
Isolating yy prepares for substitution into the quadratic equation.
2
Substitute y=2x+3y = 2x + 3 into the second equation y=x2+3y = x^2 + 3
2x+3=x2+32x + 3 = x^2 + 3
This creates a single quadratic equation in terms of xx.
3
Rearrange and factor the quadratic equation
x2−2x=0  ⟹  x(x−2)=0x^2 - 2x = 0 \implies x(x - 2) = 0
Subtracting 33 and 2x2x from both sides simplifies the equation to factorable form.
4
Solve for the non-zero (positive) value of xx
x=2x = 2
Setting x−2=0x - 2 = 0 yields x=2x = 2.

Key Concept

Solving simultaneous linear and quadratic equations by substitution
Question 22Question →

Given the simultaneous equations 2x+y=52x + y = 5 and x2+xy−y2=−5x^2 + xy - y^2 = -5, let (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) be the real solution pairs. What is the value of x1x2+y1y2x_1 x_2 + y_1 y_2?

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Answer: −5-5

Answer

-5
Substituting y=5−2xy = 5 - 2x into x2+xy−y2=−5x^2 + xy - y^2 = -5 yields −5x2+25x−20=0-5x^2 + 25x - 20 = 0, which simplifies to x2−5x+4=0x^2 - 5x + 4 = 0. The roots are x1=1x_1 = 1 and x2=4x_2 = 4, giving corresponding yy-values y1=3y_1 = 3 and y2=−3y_2 = -3. Evaluating x1x2+y1y2x_1 x_2 + y_1 y_2 gives (1)(4)+(3)(−3)=4−9=−5(1)(4) + (3)(-3) = 4 - 9 = -5.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation.
y=5−2xy = 5 - 2x
Substitution method requires isolating one variable from the linear equation.
2
Substitute y=5−2xy = 5 - 2x into the quadratic equation x2+xy−y2=−5x^2 + xy - y^2 = -5.
x2+x(5−2x)−(5−2x)2=−5x^2 + x(5 - 2x) - (5 - 2x)^2 = -5
Form a single quadratic equation in terms of xx.
3
Expand and simplify the quadratic equation.
x2+5x−2x2−(25−20x+4x2)=−5  ⟹  −5x2+25x−20=0  ⟹  x2−5x+4=0x^2 + 5x - 2x^2 - (25 - 20x + 4x^2) = -5 \implies -5x^2 + 25x - 20 = 0 \implies x^2 - 5x + 4 = 0
Convert the equation to standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Factorize the quadratic equation to find the values of xx.
(x−1)(x−4)=0  ⟹  x1=1,x2=4(x - 1)(x - 4) = 0 \implies x_1 = 1, x_2 = 4
Obtain the two roots for xx.
5
Find the corresponding yy-values using y=5−2xy = 5 - 2x.
For x1=1x_1 = 1: y1=5−2(1)=3y_1 = 5 - 2(1) = 3. For x2=4x_2 = 4: y2=5−2(4)=−3y_2 = 5 - 2(4) = -3. Solution pairs are (1,3)(1, 3) and (4,−3)(4, -3).
Calculate corresponding coordinate values for each root.
6
Compute the required expression x1x2+y1y2x_1 x_2 + y_1 y_2.
x1x2+y1y2=(1)(4)+(3)(−3)=4−9=−5x_1 x_2 + y_1 y_2 = (1)(4) + (3)(-3) = 4 - 9 = -5
Perform the final calculation requested in the stem.

Key Concept

Simultaneous Linear and Quadratic Equations
Question 23Question →

If xx and yy are real numbers satisfying the simultaneous equations x−y=4x - y = 4 and x2+y2=26x^2 + y^2 = 26, what is the value of the product xyxy?

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Answer: 5

Answer

The value of the product xyxy is 5.
Expressing xx from the linear equation yields x=y+4x = y + 4. Substituting this into x2+y2=26x^2 + y^2 = 26 gives (y+4)2+y2=26(y + 4)^2 + y^2 = 26, which expands and simplifies to y2+4y−5=0y^2 + 4y - 5 = 0. Factoring gives solutions y=1y = 1 (with x=5x = 5) and y=−5y = -5 (with x=−1x = -1). Both solution pairs (5,1)(5, 1) and (−1,−5)(-1, -5) result in the product xy=5xy = 5.

Step-by-Step Solution

1
Express xx in terms of yy using the linear equation
x=y+4x = y + 4
Isolate xx to substitute into the second equation.
2
Substitute x=y+4x = y + 4 into the quadratic equation x2+y2=26x^2 + y^2 = 26
(y+4)2+y2=26(y + 4)^2 + y^2 = 26
Reduce the system to a single quadratic equation in yy.
3
Expand and simplify into standard quadratic form
y2+4y−5=0y^2 + 4y - 5 = 0
Expanding gives 2y2+8y+16=262y^2 + 8y + 16 = 26, which simplifies by subtracting 26 and dividing by 2.
4
Solve for yy by factoring
y=1y = 1 or y=−5y = -5
The factors of y2+4y−5y^2 + 4y - 5 are (y+5)(y−1)=0(y + 5)(y - 1) = 0.
5
Compute corresponding xx values and the product xyxy
For y=1y = 1, x=5  ⟹  xy=5x = 5 \implies xy = 5; for y=−5y = -5, x=−1  ⟹  xy=5x = -1 \implies xy = 5
Substitute each yy back into x=y+4x = y + 4 and evaluate xyxy.

Key Concept

Simultaneous Linear and Quadratic Equations
Question 24Question →

Given the simultaneous equations x−y=1x - y = 1 and x2+y2=25x^2 + y^2 = 25, where xx and yy are both positive real numbers, what is the value of x+yx + y?

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Answer: 77

Answer

The value of x+yx + y is 77.
Expressing xx as y+1y + 1 and substituting into x2+y2=25x^2 + y^2 = 25 yields 2y2+2y−24=02y^2 + 2y - 24 = 0. Factoring gives y=3y = 3 (rejecting y=−4y = -4 as y>0y > 0). Substituting y=3y = 3 back gives x=4x = 4. Adding these values together yields x+y=7x + y = 7.

Step-by-Step Solution

1
Express xx in terms of yy using the linear equation.
x=y+1x = y + 1
Rearranging x−y=1x - y = 1 allows substitution into the quadratic equation.
2
Substitute x=y+1x = y + 1 into the quadratic equation x2+y2=25x^2 + y^2 = 25.
(y+1)2+y2=25  ⟹  y2+2y+1+y2=25  ⟹  2y2+2y−24=0(y + 1)^2 + y^2 = 25 \implies y^2 + 2y + 1 + y^2 = 25 \implies 2y^2 + 2y - 24 = 0
This reduces the system to a single quadratic equation in terms of yy.
3
Solve the quadratic equation for yy.
y2+y−12=0  ⟹  (y+4)(y−3)=0  ⟹  y=3y^2 + y - 12 = 0 \implies (y + 4)(y - 3) = 0 \implies y = 3 or y=−4y = -4
Dividing by 22 simplifies the equation, and factoring gives the potential roots for yy.
4
Select the positive value of yy and calculate xx and x+yx + y.
Since y>0y > 0, y=3y = 3. Then x=3+1=4x = 3 + 1 = 4, so x+y=4+3=7x + y = 4 + 3 = 7.
The question specifies that both xx and yy are positive real numbers.

Key Concept

Solving Simultaneous Linear and Quadratic Equations by Substitution
Question 25Question →

If (x,y)(x, y) is a pair of positive integers satisfying the simultaneous equations x+2y=7x + 2y = 7 and x2+y2=13x^2 + y^2 = 13, what is the value of 2x−y2x - y?

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Answer: 4

Answer

The value of 2x−y2x - y is 4.
Substituting x=7−2yx = 7 - 2y into x2+y2=13x^2 + y^2 = 13 yields the quadratic 5y2−28y+36=05y^2 - 28y + 36 = 0, which factors as (5y−18)(y−2)=0(5y - 18)(y - 2) = 0. Since xx and yy must both be positive integers, the valid solution pair is (x,y)=(3,2)(x, y) = (3, 2). Evaluating 2x−y2x - y gives 2(3)−2=42(3) - 2 = 4.

Step-by-Step Solution

1
Express xx in terms of yy using the linear equation x+2y=7x + 2y = 7.
x=7−2yx = 7 - 2y
Isolating xx allows substitution into the quadratic equation.
2
Substitute x=7−2yx = 7 - 2y into x2+y2=13x^2 + y^2 = 13 and expand.
(7−2y)2+y2=13  ⟹  49−28y+4y2+y2=13(7 - 2y)^2 + y^2 = 13 \implies 49 - 28y + 4y^2 + y^2 = 13
This reduces the system to a single quadratic equation in terms of yy.
3
Rearrange and solve the resulting quadratic equation 5y2−28y+36=05y^2 - 28y + 36 = 0.
(5y−18)(y−2)=0  ⟹  y=2(5y - 18)(y - 2) = 0 \implies y = 2 or y=3.6y = 3.6
Factoring provides the possible values for yy.
4
Identify the integer solution pair and compute the target expression 2x−y2x - y.
Since yy must be a positive integer, y=2y = 2 and x=3x = 3. Therefore, 2x−y=2(3)−2=42x - y = 2(3) - 2 = 4.
The non-integer solution is rejected based on the positive integer constraint.

Key Concept

Simultaneous Linear and Quadratic Equations
Estimated Time:1m 30s
Question 26Question →

What is the sum of the yy-values of the solution pairs that satisfy the simultaneous equations y−x=2y - x = 2 and y=x2−3x+5y = x^2 - 3x + 5?

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Answer: 88

Answer

The sum of the yy-values is 8.
Rearranging the linear equation gives y=x+2y = x + 2. Equating this to the quadratic expression gives x+2=x2−3x+5x + 2 = x^2 - 3x + 5, which simplifies to x2−4x+3=0x^2 - 4x + 3 = 0. Solving for xx yields x=1x = 1 and x=3x = 3. Substituting these into y=x+2y = x + 2 gives y=3y = 3 and y=5y = 5. The sum of these yy-values is 3+5=83 + 5 = 8.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation.
y=x+2y = x + 2
Isolating yy facilitates substitution into the quadratic equation.
2
Substitute y=x+2y = x + 2 into the quadratic equation and rearrange into standard form.
x+2=x2−3x+5  ⟹  x2−4x+3=0x + 2 = x^2 - 3x + 5 \implies x^2 - 4x + 3 = 0
Setting the two expressions for yy equal to each other forms a single quadratic equation in xx.
3
Solve the quadratic equation for xx.
(x−1)(x−3)=0  ⟹  x1=1,x2=3(x - 1)(x - 3) = 0 \implies x_1 = 1, x_2 = 3
Factoring gives the xx-coordinates of the intersection points.
4
Calculate the corresponding yy-values using y=x+2y = x + 2.
y1=1+2=3y_1 = 1 + 2 = 3 and y2=3+2=5y_2 = 3 + 2 = 5
Each xx-value must be substituted into the linear equation to find its paired yy-value.
5
Find the sum of the yy-values.
y1+y2=3+5=8y_1 + y_2 = 3 + 5 = 8
The question asks for the sum of the yy-coordinates.

Key Concept

Simultaneous Linear and Quadratic Equations

Alternative Method

Using Vieta's formulas: once the quadratic equation for xx is found as x2−4x+3=0x^2 - 4x + 3 = 0, we know x1+x2=4x_1 + x_2 = 4. Since y1=x1+2y_1 = x_1 + 2 and y2=x2+2y_2 = x_2 + 2, the sum of the yy-values is y1+y2=(x1+2)+(x2+2)=(x1+x2)+4=4+4=8y_1 + y_2 = (x_1 + 2) + (x_2 + 2) = (x_1 + x_2) + 4 = 4 + 4 = 8.
Estimated Time:1m 30s
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