# Algebra homework help!?

Can you help me solve these 2 linear systems using substitution?

1) 5x+4y=32
9x-4y=33
2) 11x-7y=-14
x-2y=-4

Substitution method :

5x + 4y = 32

Subtract 5x to each side

4y = 32 - 5x

9x - 4y = 33

Substitute 4y = 32 - 5x in the  above equation

9x - (32 - 5x) = 33

14x - 32 = 33

14x = 33 + 32

14x = 65

Divide each side by 14

x = 65 / 14

Substitute x = 65 / 14 in the equation 5x + 4y = 32

325 / 14 + 4y = 32

4y = 32 - 325 / 14

4y = (448 - 325) / 14

4y = 123 / 14

Divide each side by 4

y = 123 / 56

The x values of x , y are 65 / 14 , 123 / 56.

Substitution method :

11x -7y = -14

11x = 7y - 14

x - 2y = -4

11x - 22y = -44

Substitute 11x =7y -14 in the  above equation 11x - 22y = -44

7y - 14 - 22y = -44

-14 - 22y = -22

Divide each side by negitive one

14 + 22y = 22

Subtract 14 from each side

22y = 22 - 14

22y = 8

y = 8 / 22 =4 / 11

substitute y = 4 / 11 in the equation 11x - 7y = -14

11x - 28 / 11 = -14

Multiply each side by 11

121x - 28 = -154

121x = -154 + 28

121x = - 126

Divide each side by 121

x = -126 / 121

The values of x , y are -126 / 121 , 4 / 11.

Solution of the equations 11x - 7y = - 14 and x - 2y = - 4 is (x, y ) = (0, 2).

Substitution method

2) The equations are 11x - 7y = - 14 → (1)

and x - 2y = - 4→ (2)

Solve the second equation for x, since the coefficient of x is 1.

x = 2y - 4

Find the value of y by substituting 2y - 4 for x in equation (1).

11(2y - 4) - 7y = - 14

22y - 44 - 7y = - 14

15y  = - 14 + 44

15y = 30

y = 30/15

y = 2

Substitute 2 for y in either of two equation to find the value of x.

Second equation : x - 2(2) = - 4

x - 4 = - 4

x = - 4 + 4

x = 0

Solution (x, y ) = (0, 2)