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# 5/3x+1/3y+=-4

5/3x+1/3y=-4

-6x+5y+33

do I multiply the factions out by 3? then what?

### 1 Answer by Expert Tutors

Colby W. | Math and science tutorMath and science tutor
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5/3x+1/3y=-4

-6x+5y+33

I'm assuming you are solving a system of linear equations and that you meant to write -6x+5y=33.

Multiplying by three could be a good idea. Fractions make things messy, so our system becomes:

5x + y = -12 && -6x + 5y = 33

Our goal is to eliminate one of the variables, solve for the other and plug that value into one of the equations to get the variable we eliminated. The easiest way I see to do that is to multiply the first equation by 5 and subtract from the second equation.

(-6x + 5y  = 33)

- (25x + 5y = -60)

-31x        = 93.

Solving for x, we get x = 93/(-31) = -3. Now we plug it back into one the equations (I choose 5x + y = -12 because it's simple) and we get 5(-3) + y = -12. Solve for y: y = -12 + 15 = 3. So both x is -3 and y is 3. You can check that (x, y) = (-3, 3) satisfies both of the original equations by direct substitution.