Question

How do you solve the following system: #8x-3y=3, 4x + 2y = 10 #?

Answer 1

`(x,y)=(9/7,17/7)`

Explanation:

Given:
[1]`color(white)("XXX")8x-3y=3`
[2]`color(white)("XXX")4x+2y=10`

If we note that the `x` term in [2] is a simple multiple of that in [1],
we have an easy way to eliminate the `x` term:
Multiply [2] by `2` and subtract the result from [1]
[1]`color(white)("XXXXXXxX")8x-3y=3`
-[2]`xx2color(white)("XX")-(underline(8x+4y=20))`
[3]`color(white)("XXXXXXXXX")-7y=-17`

Dividing both sides by `(-7)`
[4]`color(white)("XXX")y=17/7`

Substituting `(17/7)` for `y` in [2]
[5]`color(white)("XXX")4x+34/7=70/7`

[6]`color(white)("XXX")4x=36/7`

[7]`color(white)("XXX")x=9/7`

Answer 2

`(x,y)=(9/7,17/7)`

Explanation:

Solve by elimination and substitution

`color(blue)(8x-3y=3`

`color(blue)(4x+2y=10`

We can eliminate `8x` in the first equation, by `4x` in the second equation if we multiply it with `-2` (with the whole equation) to get `-8x`

`rarr-2(4x+2y=10)`

Use distributive property:

`color(brown)(a(b+c)=ab+ac`

`(or)`

`color(brown)(a(b+c=x)=ab+ac=ax`

`rarr-8x-4y=-20`

Add both of the equations

`rarr(8x-3y=3)+(-8x-4y=-20)`

`rarr-7y=-17`

`color(green)(rArry=(-17)/-7=17/7`

Substitute the value of `y` to the second equation

`rarr4x+2(17/7=10)`

`rarr4x+34/7=10`

`rarr4x=10-34/7`

`rarr4x=36/7`

`color(green)(rArrx=36/7-:4=cancel36^9/7*1/cancel4^1=9/7`