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Simultaneous Linear and Quadratic Equations

26 questions

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Question 1Question →

The perimeter of a rectangular playfield is 28 m28\text{ m} and its area is 40 m240\text{ m}^2. What is the positive difference, in metres, between its length and width?

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Answer: 6

Answer

The positive difference between the length and width of the playfield is 6 metres.
Formulating the system gives x+y=14x + y = 14 and xy=40xy = 40. Substituting y=14−xy = 14 - x yields the quadratic equation x2−14x+40=0x^2 - 14x + 40 = 0, which factors into (x−10)(x−4)=0(x - 10)(x - 4) = 0. The dimensions are 10 m10\text{ m} and 4 m4\text{ m}, giving a positive difference of 10−4=6 m10 - 4 = 6\text{ m}.

Step-by-Step Solution

1
Set up linear and quadratic equations for perimeter and area
x+y=14x + y = 14 and xy=40xy = 40
Perimeter formula is 2(x+y)=282(x + y) = 28 which simplifies to x+y=14x + y = 14, and area formula is xy=40xy = 40.
2
Substitute y=14−xy = 14 - x into the quadratic area equation
x(14−x)=40  ⟹  x2−14x+40=0x(14 - x) = 40 \implies x^2 - 14x + 40 = 0
Substitution reduces the simultaneous system to a single quadratic equation in terms of xx.
3
Solve the quadratic equation by factoring
(x−10)(x−4)=0  ⟹  x=10 or x=4(x - 10)(x - 4) = 0 \implies x = 10 \text{ or } x = 4
The roots of the equation give the dimensions of the rectangle.
4
Calculate the positive difference between the two dimensions
10 - 4 = 6
Subtract the smaller dimension from the larger dimension.

Key Concept

Solving word problems involving simultaneous linear and quadratic equations
Estimated Time:1m 30s
Question 2Question →

Solve the simultaneous equations y=2x+1y = 2x + 1 and y=x2−2y = x^2 - 2. Which of the following represents the complete set of solution pairs (x,y)(x, y)?

Show answer & explanation

Answer: (3,7)(3, 7) and (−1,−1)(-1, -1)

Answer

The complete set of solution pairs (x,y)(x, y) is (3,7)(3, 7) and (−1,−1)(-1, -1).
Equating 2x+1=x2−22x + 1 = x^2 - 2 yields x2−2x−3=0x^2 - 2x - 3 = 0. Factoring gives (x−3)(x+1)=0(x - 3)(x + 1) = 0, leading to x=3x = 3 or x=−1x = -1. Substituting these xx-values into y=2x+1y = 2x + 1 gives y=7y = 7 for x=3x = 3, and y=−1y = -1 for x=−1x = -1. Thus, the solution pairs are (3,7)(3, 7) and (−1,−1)(-1, -1).

Step-by-Step Solution

1
Equate the linear expression for yy to the quadratic expression for yy.
2x+1=x2−22x + 1 = x^2 - 2
Since both expressions equal yy, setting them equal eliminates yy.
2
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−2x−3=0x^2 - 2x - 3 = 0
Subtract 2x2x and 11 from both sides.
3
Factor the quadratic equation to find the values of xx.
(x−3)(x+1)=0  ⟹  x=3 or x=−1(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1
Determine two numbers that multiply to −3-3 and add to −2-2.
4
Substitute each xx-value back into the linear equation y=2x+1y = 2x + 1 to find the corresponding yy-value.
For x=3x = 3, y=2(3)+1=7y = 2(3) + 1 = 7. For x=−1x = -1, y=2(−1)+1=−1y = 2(-1) + 1 = -1.
Calculate the exact coordinate pairs (x,y)(x, y) that satisfy both equations.

Key Concept

Solving simultaneous linear and quadratic equations using algebraic substitution.
Question 3Question →

If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the real solution pairs to the simultaneous equations x−2y=1x - 2y = 1 and x2−xy+y2=7x^2 - xy + y^2 = 7, what is the value of x1x2+y1y2x_1 x_2 + y_1 y_2?

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Answer: −11-11

Answer

The value of x1x2+y1y2x_1 x_2 + y_1 y_2 is −11-11.
Substituting x=2y+1x = 2y + 1 into x2−xy+y2=7x^2 - xy + y^2 = 7 gives 3y2+3y−6=03y^2 + 3y - 6 = 0, which reduces to y2+y−2=0y^2 + y - 2 = 0. The roots are y=1y = 1 and y=−2y = -2. Substituting back into the linear expression yields corresponding values x=3x = 3 and x=−3x = -3. The two solution pairs are (3,1)(3, 1) and (−3,−2)(-3, -2). Calculating x1x2+y1y2=(3)(−3)+(1)(−2)x_1 x_2 + y_1 y_2 = (3)(-3) + (1)(-2) yields −11-11.

Step-by-Step Solution

1
Express xx in terms of yy from the linear equation.
x=2y+1x = 2y + 1
Isolating variable xx allows direct substitution into the quadratic equation.
2
Substitute x=2y+1x = 2y + 1 into the quadratic equation x2−xy+y2=7x^2 - xy + y^2 = 7.
(2y+1)2−(2y+1)y+y2=7  ⟹  3y2+3y−6=0(2y + 1)^2 - (2y + 1)y + y^2 = 7 \implies 3y^2 + 3y - 6 = 0
This simplifies the system to a single quadratic equation in yy.
3
Solve the quadratic equation 3y2+3y−6=03y^2 + 3y - 6 = 0.
y2+y−2=0  ⟹  (y+2)(y−1)=0y^2 + y - 2 = 0 \implies (y + 2)(y - 1) = 0, giving y1=1y_1 = 1 and y2=−2y_2 = -2
Factoring determines the two possible yy-coordinates.
4
Determine corresponding xx-coordinates for each yy-value.
For y1=1y_1 = 1, x1=2(1)+1=3  ⟹  (3,1)x_1 = 2(1) + 1 = 3 \implies (3, 1). For y2=−2y_2 = -2, x2=2(−2)+1=−3  ⟹  (−3,−2)x_2 = 2(-2) + 1 = -3 \implies (-3, -2).
Substituting each yy into x=2y+1x = 2y + 1 yields the complete solution pairs.
5
Compute x1x2+y1y2x_1 x_2 + y_1 y_2.
(3)(−3)+(1)(−2)=−9−2=−11(3)(-3) + (1)(-2) = -9 - 2 = -11
Evaluates the targeted algebraic expression.

Key Concept

Solving simultaneous linear and quadratic equations using algebraic substitution.
Question 4Question →

What are the values of yy for the real solution pairs (x,y)(x, y) that satisfy the simultaneous equations y−3x=2y - 3x = 2 and y=x2−x+5y = x^2 - x + 5?

Show answer & explanation

Answer: 55 or 1111

Answer

55 or 1111
Substituting y=3x+2y = 3x + 2 into the quadratic equation y=x2−x+5y = x^2 - x + 5 gives 3x+2=x2−x+53x + 2 = x^2 - x + 5, which simplifies to x2−4x+3=0x^2 - 4x + 3 = 0. Solving this yields x=1x = 1 and x=3x = 3. Substituting these into the linear equation gives y=3(1)+2=5y = 3(1) + 2 = 5 and y=3(3)+2=11y = 3(3) + 2 = 11. Therefore, the possible values of yy are 55 or 1111.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation
y=3x+2y = 3x + 2
Linear equations can easily be substituted into quadratic equations.
2
Equate the linear expression for yy to the quadratic equation
3x+2=x2−x+53x + 2 = x^2 - x + 5
Both expressions represent yy.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x2−4x+3=0x^2 - 4x + 3 = 0
Subtract 3x3x and 22 from both sides.
4
Factorize and solve for xx
(x−1)(x−3)=0  ⟹  x=1 or x=3(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3
Finding the roots of the quadratic equation gives the xx-coordinates of the solution pairs.
5
Substitute each xx-value back into y=3x+2y = 3x + 2 to find yy
When x=1x = 1, y=3(1)+2=5y = 3(1) + 2 = 5. When x=3x = 3, y=3(3)+2=11y = 3(3) + 2 = 11.
The question specifically asks for the values of yy.

Key Concept

Simultaneous Linear and Quadratic Equations
Estimated Time:1m 30s
Question 5Question →

Find the positive value of xx that satisfies the simultaneous equations y=x+2y = x + 2 and y=x2−4y = x^2 - 4.

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Answer: 3

Answer

The positive value of xx is 3.
Equating the linear equation y=x+2y = x + 2 and the quadratic equation y=x2−4y = x^2 - 4 yields x2−x−6=0x^2 - x - 6 = 0. Factorizing this quadratic equation gives (x−3)(x+2)=0(x - 3)(x + 2) = 0, which yields roots x=3x = 3 and x=−2x = -2. Selecting the positive value gives 3.

Step-by-Step Solution

1
Equate the linear and quadratic equations
x+2=x2−4x + 2 = x^2 - 4
Since both expressions are equal to yy, set them equal to each other to solve for xx.
2
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x2−x−6=0x^2 - x - 6 = 0
Subtract xx and 22 from both sides of the equation.
3
Factorize the quadratic expression
(x−3)(x+2)=0(x - 3)(x + 2) = 0
Find two factors of −6-6 that add up to −1-1, which are −3-3 and 22.
4
Determine the roots and select the positive value
x=3x = 3
Setting each factor to zero gives x=3x = 3 or x=−2x = -2. Selecting the positive root yields 33.

Key Concept

Solving simultaneous linear and quadratic equations by substitution
Question 6Question →

Given the simultaneous equations x+2y=7x + 2y = 7 and x2+3xy+y2=19x^2 + 3xy + y^2 = 19, find the product of all possible values of yy that satisfy the system.

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Answer: -30

Answer

The product of all possible values of y that satisfy the system is -30.
Rearranging the linear equation gives x = 7 - 2y. Substituting this into the quadratic equation x^2 + 3xy + y^2 = 19 produces (7 - 2y)^2 + 3(7 - 2y)y + y^2 = 19. Expanding and combining like terms yields y^2 + 7y - 30 = 0. Solving for y gives y = 3 and y = -10. Multiplying these values together gives a product of -30.

Step-by-Step Solution

1
Isolate x in the linear equation
x = 7 - 2y
Expressing one variable in terms of the other enables substitution into the quadratic equation.
2
Substitute x into the quadratic equation and expand
(7 - 2y)^2 + 3(7 - 2y)y + y^2 = 19
This reduces the system to a single quadratic equation in terms of y.
3
Simplify the resulting expression into standard quadratic form
y^2 + 7y - 30 = 0
Expanding yields (49 - 28y + 4y^2) + (21y - 6y^2) + y^2 = 19, which reduces to y^2 + 7y - 30 = 0.
4
Calculate the product of the roots of y
y_1 * y_2 = -30
By Vieta's formulas, the product of roots for y^2 + ay + b = 0 is b/1 = -30 (or factoring gives y = 3 and y = -10, with product 3 * (-10) = -30).

Key Concept

Solving simultaneous linear and quadratic equations via substitution and applying quadratic root properties
Estimated Time:2m 0s
Question 7Question →

If (x,y)(x, y) satisfies the simultaneous equations x+2y=5x + 2y = 5 and x2+y2=10x^2 + y^2 = 10, what is the positive value of xx?

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Answer: 3

Answer

The positive value of xx is 3.
Isolating xx in the linear equation gives x=5−2yx = 5 - 2y. Substituting this expression into x2+y2=10x^2 + y^2 = 10 yields (5−2y)2+y2=10(5 - 2y)^2 + y^2 = 10. Expanding gives 25−20y+4y2+y2=10  ⟹  5y2−20y+15=025 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0. Dividing all terms by 55 produces y2−4y+3=0y^2 - 4y + 3 = 0, which factors as (y−1)(y−3)=0(y - 1)(y - 3) = 0, so y=1y = 1 or y=3y = 3. Substituting these into x=5−2yx = 5 - 2y gives x=3x = 3 when y=1y = 1 and x=−1x = -1 when y=3y = 3. The positive value of xx is 3.

Step-by-Step Solution

1
Express xx from the linear equation
x=5−2yx = 5 - 2y
Isolating xx allows substitution into the quadratic equation.
2
Substitute into the quadratic equation
(5−2y)2+y2=10(5 - 2y)^2 + y^2 = 10
Eliminates variable xx to create a single-variable equation in yy.
3
Expand and simplify
5y2−20y+15=0  ⟹  y2−4y+3=05y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0
Transforms the equation into standard quadratic form for easy factorization.
4
Solve for yy
y=1 or y=3y = 1 \text{ or } y = 3
Factoring (y−1)(y−3)=0(y - 1)(y - 3) = 0 yields the two possible values for yy.
5
Determine corresponding xx values and select the positive one
x=3x = 3 (from y=1y = 1)
Evaluating x=5−2yx = 5 - 2y gives x=3x = 3 and x=−1x = -1; the positive result requested is 3.

Key Concept

Solving simultaneous linear and quadratic equations by substitution
Estimated Time:1m 30s
Question 8Question →

If 2x+y=72x + y = 7 and x2+xy=6x^2 + xy = 6, what are the possible values of xx?

Show answer & explanation

Answer: 11 or 66

Answer

The possible values of xx are 11 or 66.
From the linear equation 2x+y=72x + y = 7, we get y=7−2xy = 7 - 2x. Substituting this into x2+xy=6x^2 + xy = 6 gives x2+x(7−2x)=6x^2 + x(7 - 2x) = 6, which simplifies to −x2+7x=6-x^2 + 7x = 6, or x2−7x+6=0x^2 - 7x + 6 = 0. Factoring gives (x−1)(x−6)=0(x - 1)(x - 6) = 0, leading to x=1x = 1 or x=6x = 6.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation.
y=7−2xy = 7 - 2x
Isolation of one variable allows substitution into the non-linear equation.
2
Substitute y=7−2xy = 7 - 2x into the second equation x2+xy=6x^2 + xy = 6.
x2+x(7−2x)=6  ⟹  x2+7x−2x2=6  ⟹  −x2+7x−6=0x^2 + x(7 - 2x) = 6 \implies x^2 + 7x - 2x^2 = 6 \implies -x^2 + 7x - 6 = 0
This reduces the system to a single quadratic equation in xx.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−7x+6=0x^2 - 7x + 6 = 0
Standard quadratic form enables easy factorization.
4
Factor the quadratic equation and solve for xx.
(x−1)(x−6)=0  ⟹  x=1 or x=6(x - 1)(x - 6) = 0 \implies x = 1 \text{ or } x = 6
Setting each factor to zero yields the values of xx.

Key Concept

Solving simultaneous linear and quadratic equations by substitution
Question 9Question →

If (x,y)(x, y) satisfies the simultaneous equations x−y=1x - y = 1 and x2+y2=25x^2 + y^2 = 25, what is the value of the product xyxy?

Show answer & explanation

Answer: 12

Answer

12
Expanding (x−y)2(x - y)^2 yields x2−2xy+y2x^2 - 2xy + y^2. Substituting the given values x−y=1x - y = 1 and x2+y2=25x^2 + y^2 = 25 into this identity gives 1=25−2xy1 = 25 - 2xy. Rearranging yields 2xy=242xy = 24, which solves to xy=12xy = 12. Alternatively, solving by substitution gives solution pairs (4,3)(4, 3) and (−3,−4)(-3, -4), both yielding a product of 1212.

Step-by-Step Solution

1
Apply the algebraic expansion identity
(x−y)2=x2+y2−2xy(x - y)^2 = x^2 + y^2 - 2xy
Connects the difference of terms, the sum of their squares, and their product.
2
Substitute the values given in the system of equations
12=25−2xy1^2 = 25 - 2xy
Replaces x−yx - y with 1 and x2+y2x^2 + y^2 with 25.
3
Isolate and calculate the product xyxy
2xy=24  ⟹  xy=122xy = 24 \implies xy = 12
Simplifies 1=25−2xy1 = 25 - 2xy to find the exact numerical value of xyxy.

Key Concept

Simultaneous Linear and Quadratic Equations
Question 10Question →

Consider the system of simultaneous equations x−3y=2x - 3y = 2 and x2−2xy−4y2=19x^2 - 2xy - 4y^2 = 19. What is the sum of the xx-values of the real solution pairs (x,y)(x, y)?

Show answer & explanation

Answer: 2828

Answer

The sum of the xx-values of the real solution pairs is 2828.
Rearranging the linear equation gives x=3y+2x = 3y + 2. Substituting this into x2−2xy−4y2=19x^2 - 2xy - 4y^2 = 19 results in (3y+2)2−2(3y+2)y−4y2=19(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19, which simplifies to y2−8y+15=0y^2 - 8y + 15 = 0. The roots are y=5y = 5 and y=3y = 3. Substituting these back into x=3y+2x = 3y + 2 yields x=17x = 17 and x=11x = 11. Their sum is 17+11=2817 + 11 = 28.

Step-by-Step Solution

1
Express xx in terms of yy using the linear equation.
x=3y+2x = 3y + 2
Isolation of xx allows direct substitution into the quadratic equation.
2
Substitute x=3y+2x = 3y + 2 into the quadratic equation x2−2xy−4y2=19x^2 - 2xy - 4y^2 = 19.
(3y+2)2−2(3y+2)y−4y2=19(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19
This reduces the non-linear system to a single quadratic equation in terms of yy.
3
Expand and simplify the quadratic equation.
(9y2+12y+4)−(6y2+4y)−4y2=19  ⟹  −y2+8y+4=19  ⟹  y2−8y+15=0(9y^2 + 12y + 4) - (6y^2 + 4y) - 4y^2 = 19 \implies -y^2 + 8y + 4 = 19 \implies y^2 - 8y + 15 = 0
Putting the quadratic expression into standard form ay2+by+c=0ay^2 + by + c = 0 facilitates finding its roots.
4
Solve the quadratic equation y2−8y+15=0y^2 - 8y + 15 = 0 for yy.
(y−5)(y−3)=0  ⟹  y1=5,y2=3(y - 5)(y - 3) = 0 \implies y_1 = 5, y_2 = 3
Factoring determines the ordinate values for the solution pairs.
5
Calculate the corresponding xx-values and find their sum.
For y1=5y_1 = 5: x1=3(5)+2=17x_1 = 3(5) + 2 = 17.
For y2=3y_2 = 3: x2=3(3)+2=11x_2 = 3(3) + 2 = 11.
Sum = 17+11=2817 + 11 = 28.
Plugging the yy-values back into x=3y+2x = 3y + 2 gives the abscissas, which are added to answer the question.

Key Concept

Solving simultaneous linear and quadratic equations by algebraic substitution
Estimated Time:2m 0s
Question 11Question →

What is the positive value of yy that satisfies the simultaneous equations x−y=2x - y = 2 and x2−y2=12x^2 - y^2 = 12?

Show answer & explanation

Answer: 2

Answer

The positive value of y is 2.
Factoring x2−y2x^2 - y^2 as (x−y)(x+y)(x - y)(x + y) gives 2(x+y)=122(x + y) = 12, leading to x+y=6x + y = 6. Subtracting x−y=2x - y = 2 from x+y=6x + y = 6 isolates 2y=42y = 4, giving y=2y = 2.

Step-by-Step Solution

1
Factorize the quadratic difference of squares equation.
x2−y2=(x−y)(x+y)=12x^2 - y^2 = (x - y)(x + y) = 12
Difference of two squares identity allows linear substitution.
2
Substitute x−y=2x - y = 2 into the factorized expression.
2(x+y)=12  ⟹  x+y=62(x + y) = 12 \implies x + y = 6
Simplifies the quadratic system into a second linear equation.
3
Solve the system of linear equations x+y=6x + y = 6 and x−y=2x - y = 2 for yy.
(x+y)−(x−y)=6−2  ⟹  2y=4  ⟹  y=2(x + y) - (x - y) = 6 - 2 \implies 2y = 4 \implies y = 2
Subtracting the two linear equations eliminates xx and directly isolates yy.

Key Concept

Simultaneous Linear and Quadratic Equations via Difference of Squares Substitution
Estimated Time:45s
Question 12Question →

The line y=2x−1y = 2x - 1 intersects the curve y=x2−4x+4y = x^2 - 4x + 4 at two distinct points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). Calculate the sum of the yy-coordinates of these two points of intersection, y1+y2y_1 + y_2.

Show answer & explanation

Answer: 10

Answer

The sum of the yy-coordinates of the points of intersection is 10.
Equating y=2x−1y = 2x - 1 and y=x2−4x+4y = x^2 - 4x + 4 yields x2−6x+5=0x^2 - 6x + 5 = 0, whose solutions are x=1x = 1 and x=5x = 5. Substituting these values into y=2x−1y = 2x - 1 gives y1=1y_1 = 1 and y2=9y_2 = 9. The sum y1+y2=1+9=10y_1 + y_2 = 1 + 9 = 10.

Step-by-Step Solution

1
Equate the linear and quadratic equations to eliminate yy.
x2−6x+5=0x^2 - 6x + 5 = 0
Setting 2x−1=x2−4x+42x - 1 = x^2 - 4x + 4 allows finding the xx-coordinates of the intersection points.
2
Solve the quadratic equation for xx.
x1=1x_1 = 1 and x2=5x_2 = 5
Factoring (x−1)(x−5)=0(x - 1)(x - 5) = 0 yields the two xx-values.
3
Determine the corresponding yy-values using y=2x−1y = 2x - 1.
y1=1y_1 = 1 and y2=9y_2 = 9
Substituting x=1x = 1 yields y=1y = 1, and substituting x=5x = 5 yields y=9y = 9.
4
Calculate the sum of the yy-coordinates.
10
Adding y1+y2=1+9=10y_1 + y_2 = 1 + 9 = 10.

Key Concept

Solving simultaneous linear and quadratic equations to find coordinates of intersection
Question 13Question →

What are the values of yy that satisfy the simultaneous equations 3x+y=103x + y = 10 and x2+y=14x^2 + y = 14?

Show answer & explanation

Answer: −2-2 or 1313

Answer

The values of yy that satisfy the simultaneous equations are −2-2 or 1313.
Rearranging the linear equation gives y=10−3xy = 10 - 3x. Substituting this into x2+y=14x^2 + y = 14 yields x2+10−3x=14x^2 + 10 - 3x = 14, which simplifies to x2−3x−4=0x^2 - 3x - 4 = 0. Factoring gives (x−4)(x+1)=0(x - 4)(x + 1) = 0, so x=4x = 4 or x=−1x = -1. Substituting x=4x = 4 into y=10−3xy = 10 - 3x gives y=−2y = -2, and substituting x=−1x = -1 gives y=13y = 13. Thus, the required values of yy are −2-2 or 1313.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation
y=10−3xy = 10 - 3x
Isolation of variable yy allows straightforward substitution into the quadratic equation.
2
Substitute y=10−3xy = 10 - 3x into the quadratic equation x2+y=14x^2 + y = 14
x2+(10−3x)=14  ⟹  x2−3x−4=0x^2 + (10 - 3x) = 14 \implies x^2 - 3x - 4 = 0
Creates a quadratic equation in terms of a single variable xx.
3
Solve the quadratic equation x2−3x−4=0x^2 - 3x - 4 = 0 for xx
(x−4)(x+1)=0  ⟹  x=4 or x=−1(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1
Factoring determines the xx-coordinates of the solution points.
4
Substitute each xx-value back into y=10−3xy = 10 - 3x to find the corresponding yy-values
For x=4x = 4: y=10−3(4)=−2y = 10 - 3(4) = -2; for x=−1x = -1: y=10−3(−1)=13y = 10 - 3(-1) = 13
Evaluates the exact values of yy requested by the question.

Key Concept

Simultaneous Linear and Quadratic Equations
Question 14Question →

If aa and bb are real numbers satisfying the simultaneous equations a+b=10a + b = 10 and a2−b2=40a^2 - b^2 = 40, what is the value of aa?

Show answer & explanation

Answer: 7

Answer

The value of aa is 7.
Using the difference of squares identity, a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b). Substituting a+b=10a+b = 10 into a2−b2=40a^2 - b^2 = 40 gives 10(a−b)=4010(a-b) = 40, which simplifies to a−b=4a-b = 4. Adding the two linear equations a+b=10a+b = 10 and a−b=4a-b = 4 eliminates bb, giving 2a=142a = 14, hence a=7a = 7.

Step-by-Step Solution

1
Factorize the quadratic expression a2−b2a^2 - b^2.
(a+b)(a−b)=40(a + b)(a - b) = 40
Apply the difference of two squares identity.
2
Substitute a+b=10a + b = 10 into the factorized equation.
10(a−b)=40  ⟹  a−b=410(a - b) = 40 \implies a - b = 4
Simplifying yields a second linear equation.
3
Solve the system of linear equations a+b=10a + b = 10 and a−b=4a - b = 4 for aa.
(a+b)+(a−b)=10+4  ⟹  2a=14  ⟹  a=7(a + b) + (a - b) = 10 + 4 \implies 2a = 14 \implies a = 7
Adding the two equations eliminates bb directly.

Key Concept

Simultaneous linear and quadratic equations involving difference of squares
Question 15Question →

Given the simultaneous equations 2x−y=42x - y = 4 and x2+y2=13x^2 + y^2 = 13, where both xx and yy are positive real numbers, calculate the value of x+yx + y.

Show answer & explanation

Answer: 5

Answer

The correct value of x+yx + y is 5.
Substituting y=2x−4y = 2x - 4 into x2+y2=13x^2 + y^2 = 13 gives 5x2−16x+3=05x^2 - 16x + 3 = 0, which yields x=3x = 3 or x=0.2x = 0.2. The corresponding yy-values are y=2y = 2 and y=−3.6y = -3.6. Since both xx and yy must be positive, the valid pair is (3,2)(3, 2), giving x+y=3+2=5x + y = 3 + 2 = 5.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation 2x−y=42x - y = 4.
y=2x−4y = 2x - 4
Substitution is the standard method for solving simultaneous linear and quadratic equations.
2
Substitute y=2x−4y = 2x - 4 into the quadratic equation x2+y2=13x^2 + y^2 = 13.
x2+(2x−4)2=13x^2 + (2x - 4)^2 = 13
This reduces the system to a single quadratic equation in one variable.
3
Expand and collect like terms.
5x2−16x+3=05x^2 - 16x + 3 = 0
Expanding (2x−4)2=4x2−16x+16(2x - 4)^2 = 4x^2 - 16x + 16 and subtracting 13 puts the quadratic in standard form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation 5x2−16x+3=05x^2 - 16x + 3 = 0 for xx.
x=3x = 3 or x=0.2x = 0.2
Factoring (5x−1)(x−3)=0(5x - 1)(x - 3) = 0 yields two real solutions for xx.
5
Find corresponding values of yy and apply the positivity constraint x>0x > 0 and y>0y > 0.
x=3,y=2x = 3, y = 2
When x=0.2x = 0.2, y=−3.6y = -3.6, which is not positive. Hence, (3,2)(3, 2) is the only valid solution pair.
6
Calculate x+yx + y.
5
3+2=53 + 2 = 5.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Estimated Time:1m 30s
Question 16Question →

What is the positive value of xx that satisfies the simultaneous equations y−x=1y - x = 1 and y=x2−5y = x^2 - 5?

Show answer & explanation

Answer: 33

Answer

The positive value of xx is 33.
Substituting y=x+1y = x + 1 from the linear equation into the quadratic equation gives x+1=x2−5x + 1 = x^2 - 5. Rearranging terms yields x2−x−6=0x^2 - x - 6 = 0, which factors as (x−3)(x+2)=0(x - 3)(x + 2) = 0. The roots are x=3x = 3 and x=−2x = -2. Selecting the positive solution gives 33.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation.
y=x+1y = x + 1
Isolating yy simplifies substitution into the quadratic equation.
2
Substitute y=x+1y = x + 1 into the quadratic equation y=x2−5y = x^2 - 5.
x+1=x2−5x + 1 = x^2 - 5
This produces a quadratic equation in one variable, xx.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−x−6=0x^2 - x - 6 = 0
Standard quadratic form allows direct factorization.
4
Factorize the quadratic equation.
(x−3)(x+2)=0(x - 3)(x + 2) = 0
Factoring determines the roots of the equation.
5
Solve for xx and identify the positive value.
x=3x = 3 or x=−2x = -2; positive value is 33.
The question specifies the positive real solution for xx.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Question 17Question →

If xx and yy satisfy the simultaneous equations x−2y=1x - 2y = 1 and x2−xy−y2=5x^2 - xy - y^2 = 5, what is the positive value of xx?

Show answer & explanation

Answer: 3

Answer

The positive value of xx is 3.
Substituting x=2y+1x = 2y + 1 into the non-linear equation yields (2y+1)2−(2y+1)y−y2=5(2y + 1)^2 - (2y + 1)y - y^2 = 5, which simplifies to y2+3y−4=0y^2 + 3y - 4 = 0. Solving this gives y=1y = 1 or y=−4y = -4. Substituting y=1y = 1 into x=2y+1x = 2y + 1 gives x=3x = 3, which is the positive value of xx.

Step-by-Step Solution

1
Make xx the subject of the linear equation
x=2y+1x = 2y + 1
Substitution is the standard method for solving linear-quadratic simultaneous systems.
2
Substitute x=2y+1x = 2y + 1 into x2−xy−y2=5x^2 - xy - y^2 = 5
(2y+1)2−(2y+1)y−y2=5(2y + 1)^2 - (2y + 1)y - y^2 = 5
Eliminates xx to create a single quadratic equation in terms of yy.
3
Expand and simplify to standard quadratic form
y2+3y−4=0y^2 + 3y - 4 = 0
Simplifying algebraic expressions allows factorization.
4
Solve for yy by factoring
y=1y = 1 or y=−4y = -4
Factors of −4-4 that sum to 33 are +4+4 and −1-1.
5
Calculate corresponding values for xx
x=3x = 3 when y=1y = 1, and x=−7x = -7 when y=−4y = -4
Substitute yy back into the linear expression for xx.
6
Select the positive value of xx
3
The question explicitly asks for the positive value of xx.

Key Concept

Simultaneous Linear and Quadratic Equations
Question 18Question →

If (u,v)(u, v) is a pair of real numbers satisfying the simultaneous equations u−2v=1u - 2v = 1 and u2−3v2=13u^2 - 3v^2 = 13, what is the sum of all possible values of uu?

Show answer & explanation

Answer: −6-6

Answer

The sum of all possible values of uu is −6-6.
From the linear equation u−2v=1u - 2v = 1, we express uu as u=2v+1u = 2v + 1. Substituting this into the quadratic equation u2−3v2=13u^2 - 3v^2 = 13 gives (2v+1)2−3v2=13(2v + 1)^2 - 3v^2 = 13, which expands and simplifies to v2+4v−12=0v^2 + 4v - 12 = 0. Factoring gives (v+6)(v−2)=0(v + 6)(v - 2) = 0, so v=2v = 2 or v=−6v = -6. Substituting these back into u=2v+1u = 2v + 1 yields u=5u = 5 (for v=2v = 2) and u=−11u = -11 (for v=−6v = -6). The sum of all possible values of uu is 5+(−11)=−65 + (-11) = -6.

Step-by-Step Solution

1
Express uu in terms of vv using the linear equation.
u=2v+1u = 2v + 1
Isolating uu allows for straightforward substitution into the quadratic equation.
2
Substitute u=2v+1u = 2v + 1 into the quadratic equation u2−3v2=13u^2 - 3v^2 = 13 and expand.
(2v+1)2−3v2=13  ⟹  (4v2+4v+1)−3v2=13  ⟹  v2+4v−12=0(2v + 1)^2 - 3v^2 = 13 \implies (4v^2 + 4v + 1) - 3v^2 = 13 \implies v^2 + 4v - 12 = 0
This reduces the system to a single quadratic equation in terms of vv.
3
Solve the quadratic equation for vv by factorisation.
(v+6)(v−2)=0  ⟹  v=2 or v=−6(v + 6)(v - 2) = 0 \implies v = 2 \text{ or } v = -6
Finding the roots of the quadratic gives the vv-coordinates of the solution pairs.
4
Calculate the corresponding values of uu using u=2v+1u = 2v + 1.
For v=2v = 2: u=2(2)+1=5u = 2(2) + 1 = 5.
For v=−6v = -6: u=2(−6)+1=−11u = 2(-6) + 1 = -11.
Each vv value corresponds to a specific uu value in the solution pairs.
5
Compute the sum of all possible values of uu.
5+(−11)=−65 + (-11) = -6
The question asks specifically for the sum of the uu-values.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Question 19Question →

If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the real solution pairs to the simultaneous equations 3x−y=53x - y = 5 and x2+2xy−y2=7x^2 + 2xy - y^2 = 7, with x1<x2x_1 < x_2, what is the value of y2−y1y_2 - y_1?

Show answer & explanation

Answer: 18

Answer

18
Expressing yy as 3x−53x - 5 and substituting into the quadratic equation yields x2−10x+16=0x^2 - 10x + 16 = 0. Solving gives x1=2x_1 = 2 and x2=8x_2 = 8. Evaluating y=3x−5y = 3x - 5 for both values gives y1=1y_1 = 1 and y2=19y_2 = 19. The difference y2−y1=19−1=18y_2 - y_1 = 19 - 1 = 18.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation
y=3x−5y = 3x - 5
Isolation of yy facilitates substitution into the non-linear equation.
2
Substitute y=3x−5y = 3x - 5 into the quadratic equation x2+2xy−y2=7x^2 + 2xy - y^2 = 7
x2+2x(3x−5)−(3x−5)2=7x^2 + 2x(3x - 5) - (3x - 5)^2 = 7
Reduces the system to a single quadratic equation in terms of xx.
3
Expand and simplify the algebraic expression
x2+6x2−10x−(9x2−30x+25)=7  ⟹  −2x2+20x−32=0  ⟹  x2−10x+16=0x^2 + 6x^2 - 10x - (9x^2 - 30x + 25) = 7 \implies -2x^2 + 20x - 32 = 0 \implies x^2 - 10x + 16 = 0
Simplifies the equation to standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation x2−10x+16=0x^2 - 10x + 16 = 0 by factorization
(x−2)(x−8)=0  ⟹  x1=2(x - 2)(x - 8) = 0 \implies x_1 = 2 and x2=8x_2 = 8
Identifies the two roots with x1<x2x_1 < x_2 as specified by the condition.
5
Determine the corresponding yy-values using y=3x−5y = 3x - 5
y1=3(2)−5=1y_1 = 3(2) - 5 = 1 and y2=3(8)−5=19y_2 = 3(8) - 5 = 19
Obtains the complete coordinate solution pairs (2,1)(2, 1) and (8,19)(8, 19).
6
Calculate the required difference y2−y1y_2 - y_1
y2−y1=19−1=18y_2 - y_1 = 19 - 1 = 18
Computes the final required target value.

Key Concept

Solving Simultaneous Linear and Quadratic Equations by Substitution
Question 20Question →

Given the simultaneous equations 2x−y=12x - y = 1 and 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17, let (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) be the real solution pairs such that x1>x2x_1 > x_2. What is the value of 2x1+y22x_1 + y_2?

Show answer & explanation

Answer: 1

Answer

1
Substituting y=2x−1y = 2x - 1 into 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17 simplifies correctly to 4x2−4x−15=04x^2 - 4x - 15 = 0. Factorizing gives roots x=52x = \frac{5}{2} and x=−32x = -\frac{3}{2}. Given x1>x2x_1 > x_2, we set x1=52x_1 = \frac{5}{2} and x2=−32x_2 = -\frac{3}{2}. Substituting x2x_2 back into the linear equation gives y2=−4y_2 = -4. Calculating 2x1+y2=2(52)+(−4)=5−4=12x_1 + y_2 = 2\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation.
y=2x−1y = 2x - 1
Substitution method requires expressing one variable in terms of the other.
2
Substitute y=2x−1y = 2x - 1 into the quadratic equation 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17.
2x2−x(2x−1)+(2x−1)2+x−(2x−1)=172x^2 - x(2x - 1) + (2x - 1)^2 + x - (2x - 1) = 17
This converts the system into a single quadratic equation in terms of xx.
3
Expand and simplify the algebraic expression.
2x2−2x2+x+4x2−4x+1+x−2x+1=17  ⟹  4x2−4x+2=17  ⟹  4x2−4x−15=02x^2 - 2x^2 + x + 4x^2 - 4x + 1 + x - 2x + 1 = 17 \implies 4x^2 - 4x + 2 = 17 \implies 4x^2 - 4x - 15 = 0
Combining like terms reveals the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation 4x2−4x−15=04x^2 - 4x - 15 = 0 by factorization.
(2x−5)(2x+3)=0  ⟹  x=52(2x - 5)(2x + 3) = 0 \implies x = \frac{5}{2} or x=−32x = -\frac{3}{2}
Finding the two real values for xx.
5
Identify x1x_1 and x2x_2 according to x1>x2x_1 > x_2 and compute corresponding yy-values.
x1=52  ⟹  y1=2(52)−1=4x_1 = \frac{5}{2} \implies y_1 = 2\left(\frac{5}{2}\right) - 1 = 4; x2=−32  ⟹  y2=2(−32)−1=−4x_2 = -\frac{3}{2} \implies y_2 = 2\left(-\frac{3}{2}\right) - 1 = -4
Determining the complete coordinate solution pairs.
6
Evaluate the expression 2x1+y22x_1 + y_2.
2(52)+(−4)=5−4=12\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1
Answering the specific value requested in the problem statement.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Estimated Time:2m 30s
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