Question

How do you solve the following system: #x-4y=2, 5x+2y=20 #?

Answer 1

`x=42/11,y=5/11`

Explanation:

Take the first equation.

`x-4y=2`

Add `4y` to both sides.

`x=4y+2`

Since we know that `x=4y+2`, we can place `x` with `4y+2` in the other equation.

`5color(red)x+2y=20`

`5color(red)((4y+2))+2y=20`

Distribute and solve for `y`.

`20y+10+2y=20`

`22y=10`

`y=10/22=5/11`

Now that we know the value for `y`, plug it into either equation to solve for `x`. I'm choosing the first equation since it's simpler, but the same answer would come from plugging the value into the second equation.

`x-4color(blue)y=2`

`x-4color(blue)((5/11))=2`

`x-20/11=2`

`x=22/11+20/11=42/11`

Since `x=42/11` and `y=5/11`, the two lines should intersect at the point `(42/11,5/11)` or `(3.818,0.455)`.

We can check a graph:

graph{(x-4y-2)(5x+2y-20)=0 [-5.26, 10.54, -3.38, 4.516]}