Simultaneous Linear and Quadratic Equations Alıştırma Soruları — JAMB UTME

Soru 1Soru →

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

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The positive difference between the length and width of the playfield is 6 metres.

Formulating the system gives

x + y = 14

and

xy = 40

. Substituting

y = 14 - x

yields the quadratic equation

x^2 - 14x + 40 = 0

, which factors into

(x - 10)(x - 4) = 0

. The dimensions are

10\text{ m}

and

4\text{ m}

, giving a positive difference of

10 - 4 = 6\text{ m}

.

Adım Adım Çözüm

1

Set up linear and quadratic equations for perimeter and area

x + y = 14

and

xy = 40

Perimeter formula is

2(x + y) = 28

which simplifies to

x + y = 14

, and area formula is

xy = 40

.

2

Substitute

y = 14 - x

into the quadratic area equation

x(14 - x) = 40 \implies x^2 - 14x + 40 = 0

Substitution reduces the simultaneous system to a single quadratic equation in terms of

x

.

3

Solve the quadratic equation by factoring

(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.

Anahtar Kavram

Solving word problems involving simultaneous linear and quadratic equations

Tahmini Süre:1m 30s

Soru 2Soru →

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

(3, 7)

and

(-1, -1)

(-3, -5)

and

(1, 3)

(3, 5)

and

(-1, -3)

(3, 7)

and

(1, 3)

Cevabı ve açıklamayı göster

Cevap: $(3, 7)$ and $(-1, -1)$

Cevap

The complete set of solution pairs

(x, y)

is

(3, 7)

and

(-1, -1)

.

Equating

2x + 1 = x^2 - 2

yields

x^2 - 2x - 3 = 0

. Factoring gives

(x - 3)(x + 1) = 0

, leading to

x = 3

or

x = -1

. Substituting these

x

-values into

y = 2x + 1

gives

y = 7

for

x = 3

, and

y = -1

for

x = -1

. Thus, the solution pairs are

(3, 7)

and

(-1, -1)

.

Adım Adım Çözüm

1

Equate the linear expression for

y

to the quadratic expression for

y

.

2x + 1 = x^2 - 2

Since both expressions equal

y

, setting them equal eliminates

y

.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 2x - 3 = 0

Subtract

2x

and

1

from both sides.

3

Factor the quadratic equation to find the values of

x

.

(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1

Determine two numbers that multiply to

-3

and add to

-2

.

4

Substitute each

x

-value back into the linear equation

y = 2x + 1

to find the corresponding

y

-value.

For

x = 3

,

y = 2(3) + 1 = 7

. For

x = -1

,

y = 2(-1) + 1 = -1

.

Calculate the exact coordinate pairs

(x, y)

that satisfy both equations.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using algebraic substitution.

Soru 3Soru →

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

-9

-11

-7

-1

Cevabı ve açıklamayı göster

Cevap: $-11$

Cevap

The value of

x_1 x_2 + y_1 y_2

is

-11

.

Substituting

x = 2y + 1

into

x^2 - xy + y^2 = 7

gives

3y^2 + 3y - 6 = 0

, which reduces to

y^2 + y - 2 = 0

. The roots are

y = 1

and

y = -2

. Substituting back into the linear expression yields corresponding values

x = 3

and

x = -3

. The two solution pairs are

(3, 1)

and

(-3, -2)

. Calculating

x_1 x_2 + y_1 y_2 = (3)(-3) + (1)(-2)

yields

-11

.

Adım Adım Çözüm

1

Express

x

in terms of

y

from the linear equation.

x = 2y + 1

Isolating variable

x

allows direct substitution into the quadratic equation.

2

Substitute

x = 2y + 1

into the quadratic equation

x^2 - xy + y^2 = 7

.

(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

y

.

3

Solve the quadratic equation

3y^2 + 3y - 6 = 0

.

y^2 + y - 2 = 0 \implies (y + 2)(y - 1) = 0

, giving

y_1 = 1

and

y_2 = -2

Factoring determines the two possible

y

-coordinates.

4

Determine corresponding

x

-coordinates for each

y

-value.

For

y_1 = 1

,

x_1 = 2(1) + 1 = 3 \implies (3, 1)

. For

y_2 = -2

,

x_2 = 2(-2) + 1 = -3 \implies (-3, -2)

.

Substituting each

y

into

x = 2y + 1

yields the complete solution pairs.

5

Compute

x_1 x_2 + y_1 y_2

.

(3)(-3) + (1)(-2) = -9 - 2 = -11

Evaluates the targeted algebraic expression.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using algebraic substitution.

Soru 4Soru →

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

5

or

11

1

or

3

-1

or

-7

3

or

5

Cevabı ve açıklamayı göster

Cevap: $5$ or $11$

Cevap

5

or

11

Substituting

y = 3x + 2

into the quadratic equation

y = x^2 - x + 5

gives

3x + 2 = x^2 - x + 5

, which simplifies to

x^2 - 4x + 3 = 0

. Solving this yields

x = 1

and

x = 3

. Substituting these into the linear equation gives

y = 3(1) + 2 = 5

and

y = 3(3) + 2 = 11

. Therefore, the possible values of

y

are

5

or

11

.

Adım Adım Çözüm

1

Express

y

in terms of

x

from the linear equation

y = 3x + 2

Linear equations can easily be substituted into quadratic equations.

2

Equate the linear expression for

y

to the quadratic equation

3x + 2 = x^2 - x + 5

Both expressions represent

y

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

x^2 - 4x + 3 = 0

Subtract

3x

and

2

from both sides.

4

Factorize and solve for

x

(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3

Finding the roots of the quadratic equation gives the

x

-coordinates of the solution pairs.

5

Substitute each

x

-value back into

y = 3x + 2

to find

y

When

x = 1

,

y = 3(1) + 2 = 5

. When

x = 3

,

y = 3(3) + 2 = 11

.

The question specifically asks for the values of

y

.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations

Tahmini Süre:1m 30s

Soru 5Soru →

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

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The positive value of

x

is 3.

Equating the linear equation

y = x + 2

and the quadratic equation

y = x^2 - 4

yields

x^2 - x - 6 = 0

. Factorizing this quadratic equation gives

(x - 3)(x + 2) = 0

, which yields roots

x = 3

and

x = -2

. Selecting the positive value gives 3.

Adım Adım Çözüm

1

Equate the linear and quadratic equations

x + 2 = x^2 - 4

Since both expressions are equal to

y

, set them equal to each other to solve for

x

.

2

Rearrange into standard quadratic form

ax^2 + bx + c = 0

x^2 - x - 6 = 0

Subtract

x

and

2

from both sides of the equation.

3

Factorize the quadratic expression

(x - 3)(x + 2) = 0

Find two factors of

-6

that add up to

-1

, which are

-3

and

2

.

4

Determine the roots and select the positive value

x = 3

Setting each factor to zero gives

x = 3

or

x = -2

. Selecting the positive root yields

3

.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution

Soru 6Soru →

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

Cevabı ve açıklamayı göster

Cevap: -30

Cevap

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.

Adım Adım Çözüm

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).

Anahtar Kavram

Solving simultaneous linear and quadratic equations via substitution and applying quadratic root properties

Tahmini Süre:2m 0s

Soru 7Soru →

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

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The positive value of

x

is 3.

Isolating

x

in the linear equation gives

x = 5 - 2y

. Substituting this expression into

x^2 + y^2 = 10

yields

(5 - 2y)^2 + y^2 = 10

. Expanding gives

25 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0

. Dividing all terms by

5

produces

y^2 - 4y + 3 = 0

, which factors as

(y - 1)(y - 3) = 0

, so

y = 1

or

y = 3

. Substituting these into

x = 5 - 2y

gives

x = 3

when

y = 1

and

x = -1

when

y = 3

. The positive value of

x

is 3.

Adım Adım Çözüm

1

Express

x

from the linear equation

x = 5 - 2y

Isolating

x

allows substitution into the quadratic equation.

2

Substitute into the quadratic equation

(5 - 2y)^2 + y^2 = 10

Eliminates variable

x

to create a single-variable equation in

y

.

3

Expand and simplify

5y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0

Transforms the equation into standard quadratic form for easy factorization.

4

Solve for

y

y = 1 \text{ or } y = 3

Factoring

(y - 1)(y - 3) = 0

yields the two possible values for

y

.

5

Determine corresponding

x

values and select the positive one

x = 3

(from

y = 1

)

Evaluating

x = 5 - 2y

gives

x = 3

and

x = -1

; the positive result requested is 3.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution

Tahmini Süre:1m 30s

Soru 8Soru →

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

1

or

6

-1

or

-6

2

or

3

-2

or

-3

Cevabı ve açıklamayı göster

Cevap: $1$ or $6$

Cevap

The possible values of

x

are

1

or

6

.

From the linear equation

2x + y = 7

, we get

y = 7 - 2x

. Substituting this into

x^2 + xy = 6

gives

x^2 + x(7 - 2x) = 6

, which simplifies to

-x^2 + 7x = 6

, or

x^2 - 7x + 6 = 0

. Factoring gives

(x - 1)(x - 6) = 0

, leading to

x = 1

or

x = 6

.

Adım Adım Çözüm

1

Express

y

in terms of

x

using the linear equation.

y = 7 - 2x

Isolation of one variable allows substitution into the non-linear equation.

2

Substitute

y = 7 - 2x

into the second equation

x^2 + xy = 6

.

x^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

x

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 7x + 6 = 0

Standard quadratic form enables easy factorization.

4

Factor the quadratic equation and solve for

x

.

(x - 1)(x - 6) = 0 \implies x = 1 \text{ or } x = 6

Setting each factor to zero yields the values of

x

.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution

Soru 9Soru →

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

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

Expanding

(x - y)^2

yields

x^2 - 2xy + y^2

. Substituting the given values

x - y = 1

and

x^2 + y^2 = 25

into this identity gives

1 = 25 - 2xy

. Rearranging yields

2xy = 24

, which solves to

xy = 12

. Alternatively, solving by substitution gives solution pairs

(4, 3)

and

(-3, -4)

, both yielding a product of

12

.

Adım Adım Çözüm

1

Apply the algebraic expansion identity

(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

1^2 = 25 - 2xy

Replaces

x - y

with 1 and

x^2 + y^2

with 25.

3

Isolate and calculate the product

xy

2xy = 24 \implies xy = 12

Simplifies

1 = 25 - 2xy

to find the exact numerical value of

xy

.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations

Soru 10Soru →

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

28

8

20

-20

Cevabı ve açıklamayı göster

Cevap: $28$

Cevap

The sum of the

x

-values of the real solution pairs is

28

.

Rearranging the linear equation gives

x = 3y + 2

. Substituting this into

x^2 - 2xy - 4y^2 = 19

results in

(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19

, which simplifies to

y^2 - 8y + 15 = 0

. The roots are

y = 5

and

y = 3

. Substituting these back into

x = 3y + 2

yields

x = 17

and

x = 11

. Their sum is

17 + 11 = 28

.

Adım Adım Çözüm

1

Express

x

in terms of

y

using the linear equation.

x = 3y + 2

Isolation of

x

allows direct substitution into the quadratic equation.

2

Substitute

x = 3y + 2

into the quadratic equation

x^2 - 2xy - 4y^2 = 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

y

.

3

Expand and simplify the quadratic equation.

(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

ay^2 + by + c = 0

facilitates finding its roots.

4

Solve the quadratic equation

y^2 - 8y + 15 = 0

for

y

.

(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

x

-values and find their sum.

For

y_1 = 5

:

x_1 = 3(5) + 2 = 17

.
For

y_2 = 3

:

x_2 = 3(3) + 2 = 11

.
Sum =

17 + 11 = 28

.

Plugging the

y

-values back into

x = 3y + 2

gives the abscissas, which are added to answer the question.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by algebraic substitution

Tahmini Süre:2m 0s

Soru 11Soru →

What is the positive value of $y$ that satisfies the simultaneous equations $x - y = 2$ and $x^2 - y^2 = 12$ ?

2

4

6

8

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The positive value of y is 2.

Factoring

x^2 - y^2

as

(x - y)(x + y)

gives

2(x + y) = 12

, leading to

x + y = 6

. Subtracting

x - y = 2

from

x + y = 6

isolates

2y = 4

, giving

y = 2

.

Adım Adım Çözüm

1

Factorize the quadratic difference of squares equation.

x^2 - y^2 = (x - y)(x + y) = 12

Difference of two squares identity allows linear substitution.

2

Substitute

x - y = 2

into the factorized expression.

2(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 = 6

and

x - y = 2

for

y

.

(x + y) - (x - y) = 6 - 2 \implies 2y = 4 \implies y = 2

Subtracting the two linear equations eliminates

x

and directly isolates

y

.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations via Difference of Squares Substitution

Tahmini Süre:45s

Soru 12Soru →

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

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The sum of the

y

-coordinates of the points of intersection is 10.

Equating

y = 2x - 1

and

y = x^2 - 4x + 4

yields

x^2 - 6x + 5 = 0

, whose solutions are

x = 1

and

x = 5

. Substituting these values into

y = 2x - 1

gives

y_1 = 1

and

y_2 = 9

. The sum

y_1 + y_2 = 1 + 9 = 10

.

Adım Adım Çözüm

1

Equate the linear and quadratic equations to eliminate

y

.

x^2 - 6x + 5 = 0

Setting

2x - 1 = x^2 - 4x + 4

allows finding the

x

-coordinates of the intersection points.

2

Solve the quadratic equation for

x

.

x_1 = 1

and

x_2 = 5

Factoring

(x - 1)(x - 5) = 0

yields the two

x

-values.

3

Determine the corresponding

y

-values using

y = 2x - 1

.

y_1 = 1

and

y_2 = 9

Substituting

x = 1

yields

y = 1

, and substituting

x = 5

yields

y = 9

.

4

Calculate the sum of the

y

-coordinates.

10

Adding

y_1 + y_2 = 1 + 9 = 10

.

Anahtar Kavram

Solving simultaneous linear and quadratic equations to find coordinates of intersection

Soru 13Soru →

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

-1

or

4

-13

or

2

-2

or

13

7

or

22

Cevabı ve açıklamayı göster

Cevap: $-2$ or $13$

Cevap

The values of

y

that satisfy the simultaneous equations are

-2

or

13

.

Rearranging the linear equation gives

y = 10 - 3x

. Substituting this into

x^2 + y = 14

yields

x^2 + 10 - 3x = 14

, which simplifies to

x^2 - 3x - 4 = 0

. Factoring gives

(x - 4)(x + 1) = 0

, so

x = 4

or

x = -1

. Substituting

x = 4

into

y = 10 - 3x

gives

y = -2

, and substituting

x = -1

gives

y = 13

. Thus, the required values of

y

are

-2

or

13

.

Adım Adım Çözüm

1

Express

y

in terms of

x

from the linear equation

y = 10 - 3x

Isolation of variable

y

allows straightforward substitution into the quadratic equation.

2

Substitute

y = 10 - 3x

into the quadratic equation

x^2 + y = 14

x^2 + (10 - 3x) = 14 \implies x^2 - 3x - 4 = 0

Creates a quadratic equation in terms of a single variable

x

.

3

Solve the quadratic equation

x^2 - 3x - 4 = 0

for

x

(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1

Factoring determines the

x

-coordinates of the solution points.

4

Substitute each

x

-value back into

y = 10 - 3x

to find the corresponding

y

-values

For

x = 4

:

y = 10 - 3(4) = -2

; for

x = -1

:

y = 10 - 3(-1) = 13

Evaluates the exact values of

y

requested by the question.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations

Soru 14Soru →

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

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

a

is 7.

Using the difference of squares identity,

a^2 - b^2 = (a+b)(a-b)

. Substituting

a+b = 10

into

a^2 - b^2 = 40

gives

10(a-b) = 40

, which simplifies to

a-b = 4

. Adding the two linear equations

a+b = 10

and

a-b = 4

eliminates

b

, giving

2a = 14

, hence

a = 7

.

Adım Adım Çözüm

1

Factorize the quadratic expression

a^2 - b^2

.

(a + b)(a - b) = 40

Apply the difference of two squares identity.

2

Substitute

a + b = 10

into the factorized equation.

10(a - b) = 40 \implies a - b = 4

Simplifying yields a second linear equation.

3

Solve the system of linear equations

a + b = 10

and

a - b = 4

for

a

.

(a + b) + (a - b) = 10 + 4 \implies 2a = 14 \implies a = 7

Adding the two equations eliminates

b

directly.

Anahtar Kavram

Simultaneous linear and quadratic equations involving difference of squares

Soru 15Soru →

Given the simultaneous equations $2x - y = 4$ and $x^2 + y^2 = 13$ , where both $x$ and $y$ are positive real numbers, calculate the value of $x + y$ .

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The correct value of

x + y

is 5.

Substituting

y = 2x - 4

into

x^2 + y^2 = 13

gives

5x^2 - 16x + 3 = 0

, which yields

x = 3

or

x = 0.2

. The corresponding

y

-values are

y = 2

and

y = -3.6

. Since both

x

and

y

must be positive, the valid pair is

(3, 2)

, giving

x + y = 3 + 2 = 5

.

Adım Adım Çözüm

1

Express

y

in terms of

x

using the linear equation

2x - y = 4

.

y = 2x - 4

Substitution is the standard method for solving simultaneous linear and quadratic equations.

2

Substitute

y = 2x - 4

into the quadratic equation

x^2 + y^2 = 13

.

x^2 + (2x - 4)^2 = 13

This reduces the system to a single quadratic equation in one variable.

3

Expand and collect like terms.

5x^2 - 16x + 3 = 0

Expanding

(2x - 4)^2 = 4x^2 - 16x + 16

and subtracting 13 puts the quadratic in standard form

ax^2 + bx + c = 0

.

4

Solve the quadratic equation

5x^2 - 16x + 3 = 0

for

x

.

x = 3

or

x = 0.2

Factoring

(5x - 1)(x - 3) = 0

yields two real solutions for

x

.

5

Find corresponding values of

y

and apply the positivity constraint

x > 0

and

y > 0

.

x = 3, y = 2

When

x = 0.2

,

y = -3.6

, which is not positive. Hence,

(3, 2)

is the only valid solution pair.

6

Calculate

x + y

.

5

3 + 2 = 5

.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using substitution

Tahmini Süre:1m 30s

Soru 16Soru →

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

2

3

4

5

Cevabı ve açıklamayı göster

Cevap: $3$

Cevap

The positive value of

x

is

3

.

Substituting

y = x + 1

from the linear equation into the quadratic equation gives

x + 1 = x^2 - 5

. Rearranging terms yields

x^2 - x - 6 = 0

, which factors as

(x - 3)(x + 2) = 0

. The roots are

x = 3

and

x = -2

. Selecting the positive solution gives

3

.

Adım Adım Çözüm

1

Express

y

in terms of

x

using the linear equation.

y = x + 1

Isolating

y

simplifies substitution into the quadratic equation.

2

Substitute

y = x + 1

into the quadratic equation

y = x^2 - 5

.

x + 1 = x^2 - 5

This produces a quadratic equation in one variable,

x

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - x - 6 = 0

Standard quadratic form allows direct factorization.

4

Factorize the quadratic equation.

(x - 3)(x + 2) = 0

Factoring determines the roots of the equation.

5

Solve for

x

and identify the positive value.

x = 3

or

x = -2

; positive value is

3

.

The question specifies the positive real solution for

x

.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using substitution

Soru 17Soru →

If $x$ and $y$ satisfy the simultaneous equations $x - 2y = 1$ and $x^2 - xy - y^2 = 5$ , what is the positive value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The positive value of

x

is 3.

Substituting

x = 2y + 1

into the non-linear equation yields

(2y + 1)^2 - (2y + 1)y - y^2 = 5

, which simplifies to

y^2 + 3y - 4 = 0

. Solving this gives

y = 1

or

y = -4

. Substituting

y = 1

into

x = 2y + 1

gives

x = 3

, which is the positive value of

x

.

Adım Adım Çözüm

1

Make

x

the subject of the linear equation

x = 2y + 1

Substitution is the standard method for solving linear-quadratic simultaneous systems.

2

Substitute

x = 2y + 1

into

x^2 - xy - y^2 = 5

(2y + 1)^2 - (2y + 1)y - y^2 = 5

Eliminates

x

to create a single quadratic equation in terms of

y

.

3

Expand and simplify to standard quadratic form

y^2 + 3y - 4 = 0

Simplifying algebraic expressions allows factorization.

4

Solve for

y

by factoring

y = 1

or

y = -4

Factors of

-4

that sum to

3

are

+4

and

-1

.

5

Calculate corresponding values for

x

x = 3

when

y = 1

, and

x = -7

when

y = -4

Substitute

y

back into the linear expression for

x

.

6

Select the positive value of

x

3

The question explicitly asks for the positive value of

x

.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations

Soru 18Soru →

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

-6

-4

6

-8

Cevabı ve açıklamayı göster

Cevap: $-6$

Cevap

The sum of all possible values of

u

is

-6

.

From the linear equation

u - 2v = 1

, we express

u

as

u = 2v + 1

. Substituting this into the quadratic equation

u^2 - 3v^2 = 13

gives

(2v + 1)^2 - 3v^2 = 13

, which expands and simplifies to

v^2 + 4v - 12 = 0

. Factoring gives

(v + 6)(v - 2) = 0

, so

v = 2

or

v = -6

. Substituting these back into

u = 2v + 1

yields

u = 5

(for

v = 2

) and

u = -11

(for

v = -6

). The sum of all possible values of

u

is

5 + (-11) = -6

.

Adım Adım Çözüm

1

Express

u

in terms of

v

using the linear equation.

u = 2v + 1

Isolating

u

allows for straightforward substitution into the quadratic equation.

2

Substitute

u = 2v + 1

into the quadratic equation

u^2 - 3v^2 = 13

and expand.

(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

v

.

3

Solve the quadratic equation for

v

by factorisation.

(v + 6)(v - 2) = 0 \implies v = 2 \text{ or } v = -6

Finding the roots of the quadratic gives the

v

-coordinates of the solution pairs.

4

Calculate the corresponding values of

u

using

u = 2v + 1

.

For

v = 2

:

u = 2(2) + 1 = 5

.
For

v = -6

:

u = 2(-6) + 1 = -11

.

Each

v

value corresponds to a specific

u

value in the solution pairs.

5

Compute the sum of all possible values of

u

.

5 + (-11) = -6

The question asks specifically for the sum of the

u

-values.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using substitution

Soru 19Soru →

If $(x_1, y_1)$ and $(x_2, y_2)$ are the real solution pairs to the simultaneous equations $3x - y = 5$ and $x^2 + 2xy - y^2 = 7$ , with $x_1 < x_2$ , what is the value of $y_2 - y_1$ ?

18

6

20

14

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

18

Expressing

y

as

3x - 5

and substituting into the quadratic equation yields

x^2 - 10x + 16 = 0

. Solving gives

x_1 = 2

and

x_2 = 8

. Evaluating

y = 3x - 5

for both values gives

y_1 = 1

and

y_2 = 19

. The difference

y_2 - y_1 = 19 - 1 = 18

.

Adım Adım Çözüm

1

Express

y

in terms of

x

from the linear equation

y = 3x - 5

Isolation of

y

facilitates substitution into the non-linear equation.

2

Substitute

y = 3x - 5

into the quadratic equation

x^2 + 2xy - y^2 = 7

x^2 + 2x(3x - 5) - (3x - 5)^2 = 7

Reduces the system to a single quadratic equation in terms of

x

.

3

Expand and simplify the algebraic expression

x^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

ax^2 + bx + c = 0

.

4

Solve the quadratic equation

x^2 - 10x + 16 = 0

by factorization

(x - 2)(x - 8) = 0 \implies x_1 = 2

and

x_2 = 8

Identifies the two roots with

x_1 < x_2

as specified by the condition.

5

Determine the corresponding

y

-values using

y = 3x - 5

y_1 = 3(2) - 5 = 1

and

y_2 = 3(8) - 5 = 19

Obtains the complete coordinate solution pairs

(2, 1)

and

(8, 19)

.

6

Calculate the required difference

y_2 - y_1

y_2 - y_1 = 19 - 1 = 18

Computes the final required target value.

Anahtar Kavram

Solving Simultaneous Linear and Quadratic Equations by Substitution

Soru 20Soru →

Given the simultaneous equations $2x - y = 1$ and $2x^2 - xy + y^2 + x - y = 17$ , let $(x_1, y_1)$ and $(x_2, y_2)$ be the real solution pairs such that $x_1 > x_2$ . What is the value of $2x_1 + y_2$ ?

1

3

9

-3

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

1

Substituting

y = 2x - 1

into

2x^2 - xy + y^2 + x - y = 17

simplifies correctly to

4x^2 - 4x - 15 = 0

. Factorizing gives roots

x = \frac{5}{2}

and

x = -\frac{3}{2}

. Given

x_1 > x_2

, we set

x_1 = \frac{5}{2}

and

x_2 = -\frac{3}{2}

. Substituting

x_2

back into the linear equation gives

y_2 = -4

. Calculating

2x_1 + y_2 = 2\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1

.

Adım Adım Çözüm

1

Express

y

in terms of

x

from the linear equation.

y = 2x - 1

Substitution method requires expressing one variable in terms of the other.

2

Substitute

y = 2x - 1

into the quadratic equation

2x^2 - xy + y^2 + x - y = 17

.

2x^2 - x(2x - 1) + (2x - 1)^2 + x - (2x - 1) = 17

This converts the system into a single quadratic equation in terms of

x

.

3

Expand and simplify the algebraic expression.

2x^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

ax^2 + bx + c = 0

.

4

Solve the quadratic equation

4x^2 - 4x - 15 = 0

by factorization.

(2x - 5)(2x + 3) = 0 \implies x = \frac{5}{2}

or

x = -\frac{3}{2}

Finding the two real values for

x

.

5

Identify

x_1

and

x_2

according to

x_1 > x_2

and compute corresponding

y

-values.

x_1 = \frac{5}{2} \implies y_1 = 2\left(\frac{5}{2}\right) - 1 = 4

;

x_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

2x_1 + y_2

.

2\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1

Answering the specific value requested in the problem statement.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using substitution

Tahmini Süre:2m 30s

Simultaneous Linear and Quadratic Equations

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