How do you solve the system #2x-3y=12# and #x=4y+1# by substitution?

Answer 1
The solutions are #y=2# and #x=9#.
The system #2x-3y=12# and #x=4y+1# need to be solved.
The solution to #x=4y+1# has already been found. Enter it into the other equation, #2x-3y=12#.
#2(4y+1)-3y=12#
Give out the #2#.
#8y+2-3y=12#

Make it simple.

#5y+2=12#
#5y=10#
By #5#, divide both sides.
#y=2#
Enter the #2# instead of the #y# in one of the initial equations.
#x=4*2+1=9#
#x=9#
To verify the answers, replace #x# and #y# with the appropriate values.
#2x-3y=12#
#2*9-3*2=12#
#18-6=12#
#12=12#
#x=4y+1#
#9=4*2+1#
#9 is equal to 9#.

Both formulas are correct!

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Answer 2

To solve the system using substitution:

  1. Substitute the expression for x from the second equation into the first equation.
  2. Solve the resulting equation for y.
  3. Once you have the value of y, substitute it back into one of the original equations to find the value of x.

The solution is x = 17 and y = 4.

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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.

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.

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.

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.

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