Question

How do you solve `2x + 3y = 12 and 2x + y = 4`?

Answer 1

See a solution process below:

Explanation:

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

`2x + y = 4`

`-color(red)(2x) + 2x + y = -color(red)(2x) + 4`

`0 + y = -2x + 4`

`y = -2x + 4`

Step 2) Substitute `(-2x + 4)` for `y` in the first equation and solve for `x`:

`2x + 3y = 12` becomes:

`2x + 3(-2x + 4) = 12`

`2x + (3 xx -2x) + (3 xx 4) = 12`

`2x + (-6x) + 12 = 12`

`2x - 6x + 12 = 12`

`(2 - 6)x + 12 = 12`

`-4x + 12 = 12`

`-4x + 12 - color(red)(12) = 12 - color(red)(12)`

`-4x + 0 = 0`

`-4x = 0`

`(-4x)/color(red)(-4) = 0/color(red)(-4)`

`(color(red)(cancel(color(black)(-4)))x)/cancel(color(red)(-4)) = 0`

`x = 0`

Step 3) Substitute `0` for `x` in the solution to the second equation at the end of Step 1 and calculate `y`:

`y = -2x + 4` becomes:

`y = (-2 xx 0) + 4`

`y = 0 + 4`

`y = 4`

The Solution Is: `x = 0` and `y = 4` or `(0, 4)`

Answer 2

`y=4 and x=0`

Explanation:

`2x+3y=12`-----------(1)

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

`(1)-(2) 2y=8`

`:.y=4`

substitute `y=4 in (1)`

`:.2x+3(4)=12`

`:.2x+12=12`

`:.2x=12-12`

`:.2x=0`

`:.x=0`

~~~~~~~~~~~~~~~

substitute`y=4 and x=0` in (2)

`:.2(0)+4=4`

`:.0+4=4`