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How do you solve the system #y=5x-7 # and #-3x-2y=-12# by substitution?

Answer 1

Since we already know that #y=5x-7#, we can simply change #y# in the other equation:

#-3x-2y=-12# #-3x-2*(5x-7)=-12# #-3x-10x+14=-12# #-13x=-12-14# (dividing each side by #14#) #-13x=-26# #x=2# (dividing each side by #-13#)

Using the first equation, we can find #y# now that we know #x#:

#y=5*2-7#, #y=10-7#, #y=3#, and #y=5x-7#

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Answer 2

Eli Assouline

To solve the system using substitution:

Substitute (y = 5x - 7) from the first equation into the second equation.
Solve the resulting equation for (x).
Once you find the value of (x), substitute it back into either of the original equations to find the corresponding value of (y).

Starting with the second equation:

[-3x - 2(5x - 7) = -12]

Now, solve for (x):

[-3x - 10x + 14 = -12]

[-13x + 14 = -12]

[-13x = -12 - 14]

[-13x = -26]

[x = \frac{-26}{-13}]

[x = 2]

Now that you have found (x = 2), substitute it back into the first equation:

[y = 5(2) - 7]

[y = 10 - 7]

[y = 3]

So, the solution to the system is (x = 2) and (y = 3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.