Question

How do you solve 6x+5y=10 and 7x-3y=24?

Answer 1

There are two common ways:

• Substitution
• Elimination

I can show both ways, but in this case I think elimination would be easier.

`6x + 5y = 10`
`7x - 3y = 24`

I got:

`(x,y) = (150/53, -74/53)`

SUBSTITUTION

Solve for one variable first.

`6x = 10 - 5y`

`=> x = 5/3 - 5/6 y`

Plug it into the second equation to solve for the other variable.

`7(5/3 - 5/6 y) - 3y = 24`

`35/3 - 35/6 y - 3y = 24`

Multiply through by `6` to make this look nicer.

`70 - 35y - 18y = 144`

`- 53y= 144 - 70`

`color(green)(y) = (144 - 70)/(-53) = color(green)(-74/53)`

Therefore, for `x` we get:

`color(green)(x) = 5/3 - 5/6(-74/53)`

`= 5/3 + 370/318 = 5/3 + 185/159`

`= 795/477 + 555/477 = 1350/477`

`= color(green)(150/53)`

So, apparently, `color(blue)((x,y) = (150/53, -74/53))`... Let's check our answer.

`6(150/53) + 5(-74/53) stackrel(?" ")(=) 10`

`900/53 - 370/53 stackrel(?" ")(=) 10 => 530/53 = 10` `color(blue)(sqrt"")`

`7(150/53) - 3(-74/53) stackrel(?" ")(=) 24`

`1050/53 + 222/53 stackrel(?" ")(=) 24 => 1272/53 = (12 cdot 106)/53 = 24 color(blue)(sqrt"")`

Yep, it's right! Wow, not nice-looking at all!

ELIMINATION

In this case we would be scaling one of the equations with fractions to eliminate a variable.

`" "3/5(6x + cancel(5y) = 10)`
`+ " "7x - cancel(3y) = 24`
`bar(" "" "" "" "" "" "" "" ")`
`" "18/5x + 7x = 30`

`(18/5 + 35/5)x = 30`

`color(green)(x) = 30/(18/5 + 35/5)`

`= 30/(53/5) = color(green)(150/53)`

And now, plug it into the second equation.

`7(150/53) - 3y = 24`

`7(150/53) - 24 = 3y`

`color(green)(y) = 7/3(150/53) - 8`

`= 7(50/53) - 424/53`

`= 350/53 - 424/53`

`= color(green)(-74/53)`

And as before, we get:

`color(blue)((x,y) = (150/53, -74/53))`

So if you get something ugly like this, it's right!