Question

How do you solve the system of equations `8x + 8y = - 8` and `8x + 3y = 17` using elimination?

Answer 1

`"The answer : "`{4,-5}

Explanation:

`8x+8y=-8" (1) "`

`8x+3y=17" (2)"`

`"we can subtract the equation (2) from (1) to eliminate 8x."`

`8x+8y-(8x+3y)=-8-17`

`"let us rearrange the equation above"`

`8x+8y-8x-3y=-25`

`cancel(8x)+8ycancel(-8x)-3y=-25`

`8y-3y=-25`

`5y=-25`

`"Both sides of equation can be divided by 5"`

`(5y)/5=-25/5`

`(cancel(5)y)/cancel(5)=-25/5`

`color(red)(y=-5)`

`"now , we can use one of equation above to solve for 'x'"`

`8x+8y=-8`

`cancel(8)x+cancel(8)y=-cancel(8)`

`x+y=-1`

`x color(red)(-5)=-1`

`x=5-1`

`x=4`

Answer 2

`(4,-5)`

Explanation:

Labelling the equations.

`color(red)(8x)+8y=-8to(1)`

`color(red)(8x)+3y=17to(2)`

Note that the terms in x are the same in both equations. Thus if we subtract them we will eliminate the x term and be left with an equation in one variable which we can solve.

`rArr(1)-(2)" term by term on both sides."`

`(8x-8x)+(8y-3y)=(-8-17)`

`rArr5y=-25`

divide both sides by 5

`(cancel(5) y)/cancel(5)=(-25)/5`

`rArry=-5`

Substitute this value into either of the 2 equations to find the corresponding value for x

`"Substituting "y=-5to(2)`

`rArr8x+(3xx-5)=17`

`rArr8x-15=17`

add 15 to both sides.

`8xcancel(-15)cancel(+15)=17+15`

`rArr8x=32`

divide both sides by 8

`(cancel(8) x)/cancel(8)=32/8`

`rArrx=4`

`color(blue)"As a check"`

Substitute the values for x and y into both equations to check they are satisfied.

`to(1): (8xx4)+(8xx-5)=32-40=-8rArr"true"`

`to(2): (8xx4)+(3xx-5)=32-15=17rArr"true"`

`rArr(4,-5)" is the solution"`

The graph illustrates the solution.

graph{(y+x+1)(y+8/3x-17/3)=0 [-14.23, 14.25, -7.12, 7.12]}