SOLVING LINEAR AND QUADRATIC EQUATIONS

Solve each system by substitution. Check your answers.

Problem 1 :

y = x² + 4x + 1

y = x + 1

Solution:

Given equations,

y = x² + 4x + 1 ---> (1)

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

Substitute y = x + 1 in equation (1), we get

x + 1 = x² + 4x + 1

x² + 4x - x + 1 - 1 = 0

x² + 3x = 0

x(x + 3) = 0

x = 0 and x = -3

So, the points of intersection are (0, 1) and (-3, -2).

Problem 2 :

y = -x² + 2x + 10

y = x + 4

Solution:

Given equations,

y = -x² + 2x + 10 ---> (1)

y = x + 4 ---> (2)

Substitute y = x + 1 in equation (1), we get

x + 4 = -x² + 2x + 10

x² - 2x + x + 4 - 10 = 0

x² - x - 6 = 0

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

x = -2 and x = 3

So, the points of intersection are (-2, 2) and (3, 7).

Problem 3 :

y = -x² + x - 1

y = -x - 1

Solution:

Given equations,

y = -x² + x - 1 ---> (1)

y = -x - 1 ---> (2)

Substitute y = -x - 1 in equation (1), we get

-x - 1 = -x² + x - 1

x² - x - x - 1 + 1 = 0

x² - 2x = 0

x(x - 2) = 0

x = 0 and x = 2

So, the points of intersection are (0, -1) and (2, -3).

Problem 4 :

y = 2x² - 3x - 1

y = x - 3

Solution:

Given equations,

y = 2x² - 3x - 1 ---> (1)

y = x - 3 ---> (2)

Substitute y = x - 3 in equation (1), we get

x - 3 = 2x² - 3x - 1

2x² - 3x - x - 1 + 3 = 0

2x² - 4x + 2 = 0

2x² - 2x - 2x + 2 = 0

2(x² - x - x + 1) = 0

2(x² - 2x + 1) = 0

2(x - 1)(x - 1) =0

x = 1 

When x = 1

y = x - 3

y = 1 - 3

y = -2

So, the points of intersection is (1, -2).

Problem 5 :

y = x² - 3x - 20

y = -x - 5

Solution:

Given equations,

y = x² - 3x - 20 ---> (1)

y = -x - 5 ---> (2)

Substitute y = -x - 5 in equation (1), we get

-x - 5 = x² - 3x - 20

x² - 3x + x - 20 + 5 = 0

x² - 2x - 15 = 0

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

x = -3 and x = 5

So, the points of intersection are (-3, -2) and (5, -10).

Problem 6 :

y = -x² - 5x - 1

y = x + 2

Solution:

Given equations,

y = -x² - 5x - 1 ---> (1)

y = x + 2 ---> (2)

Substitute y = x + 2 in equation (1), we get

x + 2 = -x² - 5x - 1 

x² + 5x + x + 1 + 2 = 0

x² + 6x + 3 = 0

So, the points of intersection are (-0.55, 1.45) and (-5.45, -3.45).