SOLUTION: I need to show this problem can be solved by substitution, can you please help me? -4x + 2y = 16 and 3x -y = 2 These are systems of linear equations

Question 247831: I need to show this problem can be solved by substitution, can you please help me? -4x + 2y = 16 and 3x -y = 2 These are systems of linear equations
Found 2 solutions by richwmiller, solver91311:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
-4x + 2y = 16
3x -y = 2
add y to both sides of equation 2
subtract 2 from both sides of equation2
y=3x-2
substitute 3x-2 in equation 1 for y
-4x + 2(3x-2) = 16
-4x+6x-4=16
add 4 to both sides
combine
2x=20
x=10
y=3x-2
y=30-2
y=28
check
-4((10)+2(28)=16
-40+56=16
16=16
ok

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

To solve by substitution, solve either one of your equations for one of the variables in terms of the other. This will give you an expression that can be substituted in the other equation. Mathematically, it doesn't matter which equation you choose or which variable you choose -- it all comes out to the same thing in the end given correctly performed arithmetic. However, sometimes it is easier to manipulate one way than another. Your problem is like this.

Compare all of your coefficients. If you try to solve either of your equations for , you will either have to divide by 4 or by 3. Either way, somewhere you will end up with a fraction to deal with. On the other hand, solving for means that you will have to divide by 2 or -1, and that means no fractions.

Let's solve the second equation for :

Add -3x to both sides:

Multiply both sides by -1:

Now you have something in terms of that you can substitute for in your other equation:

So:

Which is a single equation in a single variable that can be solved by ordinary means. Once you have solved for , you can use that value to substitute into either equation and then solve for . Write back and let me know what you obtained for a solution set for this system.

John

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