solve each system of equations using substitution

Algebra
Tutor: None Selected Time limit: 1 Day

2y - 3/4x = -10

y+1/4x=0

Nov 18th, 2014

Solve one of the equations for x:

y + 1/4x = 0

(I'm going to multiply the entire equation by 4 to get rid of the fraction):

4(y+1/4x = 0)

4y + x = 0

x = -4y

Now substitute x = -4y into the first equation:

2y - 3/4x = -10

2y - 3/4(4y) = -10

2y - 3y = -10

-y = -10

y = 10


Substitute y = 10 into the first equation.  Again, I'm going to use the equation multiplied by 4 because it's easier:

4y + x = 0

4(10) + x = 0

40 + x = 0

x = -40

Nov 18th, 2014

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