Question

How do you solve the system of equations `2x + y = 4` and `5x + 4y = 7`?

Answer 1

`(x,y)to(3,-2)`

Explanation:

`2x+y=4to(1)`

`5x+4y=7to(2)`

`"rearrange "(1)" to obtain y in terms of x"`

`rArry=4-2xto(3)`

`color(blue)"substitute "y=4-2x" in equation "(2)`

`rArr5x+4(4-2x)=7`

`rArr5x+16-8x=7`

`rArr-3x+16=7`

`"subtract 16 from both sides"`

`-3xcancel(+16)cancel(-16)=7-16`

`rArr-3x=-9`

`rArrx=(-9)/(-3)=3`

`"substitute "x=3" in equation "(3)`

`rArry=4-6=-2`

`"the point of intersection "=(3,-2)`
graph{(y-4+2x)(y+5/4x-7/4)((x-3)^2+(y+2)^2-0.04)=0 [-10, 10, -5, 5]}

Answer 2

The solution is `x=3` and `y=-2`.

Explanation:

Given (1) `2x+y=4` and (2) `5x+4y=7`.

First I'm going to solve (1) for `y`:

(3 )`y=4-2x`

Now take equation (3) and substitute into equation (2):

`5x + 4(4-2x)=7`

Distribute:

`5x+16-8x =7`

collect like terms:

`-3x+16=7`

subtract 16 from both sides:

`-3x=-9`

divide through by `-3`:

`x=3`

Now take the value of `x` and subsitute into either of the original equations. I'll use equation (1) since it's a bit simpler:

Substitute for `x`:
`2(3) + y = 4`

Multiply:
`6+y=4`

Subtract 6 from both sides:
`y=-2`

So the solution is `x=3` and `y=-2`.