Question

How do you solve the following system?: `4x -2y =3 , -3x -y = -6`

Answer 1

See a solution process below:

Explanation:

Step 1) Solve the second equation for `y`:

`-3x - y = -6`

`-3x - y + color(red)(y) + color(blue)(6) = -6 + color(blue)(6) + color(red)(y)`

`-3x - 0 + color(blue)(6) = 0 + color(red)(y)`

`-3x + 6 = y`

`y = -3x + 6`

Step 2) Substitute `(-3x + 6)` for `y` in the first equation and solve for `x`:

`4x - 2y = 3` becomes:

`4x - 2(-3x + 6) = 3`

`4x - (2 xx -3x) - (2 xx 6) = 3`

`4x - (-6x) - 12 = 3`

`4x + 6x - 12 = 3`

`(4 + 6)x - 12 = 3`

`10x - 12 = 3`

`10x - 12 + color(red)(12) = 3 + color(red)(12)`

`10x - 0 = 15`

`10x = 15`

`(10x)/color(red)(10) = 15/color(red)(10)`

`(color(red)(cancel(color(black)(10)))x)/cancel(color(red)(10)) = 15/10`

`x = 3/2`

Step 3) Substitute `3/2` for `x` in the solution to the second equation at the end of Step 1 and calculate `y`:

`y = -3x + 6` becomes:

`y = (-3 xx 3/2) + 6`

`y = (-9)/2 + 6`

`y = (-9)/2 + (2/2 xx 6)`

`y = (-9)/2 + 12/2`

`y = (-9 + 12)/2`

`y = 3/2`

The Solution Is:

`x = 3/2` and `y = 3/2`

Or

`(3/2, 3/2)`

Answer 2

`y=3/2, x=3/2`

Explanation:

`4x-2y=3-------(1)`

`-3x-y=-6-------(2)`

`(2)xx2`

`:.-6x-2y=-12------(3)`

`(1)-(3)`

`:.10x=15`

`:.x=cancel 15^3/cancel 10^2`

`:.x=3/2`

`"substitute x"=3/2 "in" (2)`

`:.-3(3/2)-y=-6`

`:.-9/2-y=-12/2`

`:.-y=-12/2+9/2`

`:.-y=-3/2`

`:.y=3/2`

~~~~~~~~~~~~~~~~
check:-

substitute `y=3/2 and x=3/2 "in" (1)`

`:.4(3/2)-2(3/2)=3`

`:.12/2-6/2=3`

`:.cancel 6^3/cancel2^1=3`

`:.3=3`