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

##### 1 Answer

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

#70 - 35y - 18y = 144#

#- 53y= 144 - 70#

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

Therefore, for

#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,

#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!