Algebra homework help!?

Question

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

Answer 1

Given linear equation are

1)5x + 4y = 32-->1

9x - 4y = 33-->2

Add equation 1 and 2

5x + 4y = 32-->1

9x - 4y = 33-->2

-----------------------

13x      = 65

13x = 65

x = 65/13

x = 5

Substitute x = 5 in equation 1

5(5)+ 4y = 32

25 + 4y = 32

4y = 32 -25

4y = 7

y = 7/4

Therefore x = 5 and y = 7/4

2)11x - 7y = -14-->1

 x - 2y = -4-->2

Multiply equation 2 with 11

(x - 2y = -4)*11

11x - 22x = -44-->2

Substarct equation 1 and 2

11x - 7y = -14-->1

11x - 22y = -44-->2

(-)     (+)      (+)

-----------------------

         15y =  30

y = 30 /15

y = 2

Substitute y = 2 in equation 2

x - 2(2) = -4

x - 4 = -4

x = 0

Therefore x = 0 and y = 2

 

 

Comments

1) Solution of the system 5x + 4y = 32 and 9x - 4y = 33 is x = 65/14 and y = 123/56.

Answer 2

Substitution method

The equations are 5x + 4y = 32 and 9x - 4y = 33

Solve the first equation for y.

5x + 4y = 32

4y = 32 - 5x

y = (32 - 5x)/4

Substitute the y value in 9x - 4y = 33.

9x - 4[(32 - 5x)/4] = 33

9x - 32 + 5x = 33

14x = 33 + 32

14x = 65

x = 65/14

Substitute x value in 5x + 4y = 32.

5(65/14) + 4y = 32

 325/14 + 4y = 32

4y = 32 - 325/14

4y = (448-325)/14

4y = 123/14

y = 123/56

Solution x = 65/14 and y = 123/56.