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## Solving systems of equations

1. Use substitution to solve the system of equations:

-x + 2y=7

x - 3y = 3

2. Solve the system of equations by the elimination method

a.  2x+ y = -17

x - y = 2

b. 3x + y = -15

x + 2y = -10

1. Using substitution to solve the first system of eqations

step 1: isolate either x or y in one of the equations - I chose to isolate x in x - 3y = 3

x - 3y = 3
+3y      +3y
x = 3 + 3y

step 2: plug that value of x into the other equation

-x + 2y = 7             (plut in your x and solve for y!)
-(3+3y) + 2y = 7     (now y is the only varialbe in your equation!)
-3 -3y +2y = 7        (distribute the - outside of the parentheses)
-3 - y = 7               (combine all of your like terms (the y's))
+3       +3
-y = 10
y = -10

step 3: plug your value of y into one of the equations and solve for x

-x + 2y = 7
-x + 2(-10) = 7
-x + (-20) = 7
-x - 20 = 7
+20    +20
-x = 27
x = -27

Step 4: plug your values of x and y back into both eqations to make sure you did your math right

y = -10
x = -27

x - 3y = 3
-27 - 3(-10) = 3
-27 - (-30) = 3
-27 + 30 = 3
3 = 3  YUP!

-x + 2y = 7
- (-27) + 2 (-10) = 7
27 + (-20) = 7
27 - 20 = 7
7 = 7   YUP!

2. Using elimination to solve system of equations

a.  step 1: line your eqations up like you're going to add them together (the first one is easy because the top eqations has +y and the bottom one has -y so you do not need to multiply either one of them to find common (but opposite) values of x and y to get them to cancel out

2x + y = -17
+  x   -  y = 2
3x +0y = -15
3x = -15
3        3
x = -5

step 2: now you know the value of x, you can plug it back into one of the equations and solve for y! The second equation is the best since there are no coefficients of x or y

x - y = 2
-5 - y = 2
+5      +5
-y = 7
y = -7

Step 3: plug x and y into eqations to check your answers!

2x + y = -17
2(-5) + (-7) = -17
-10 + (-7) = =17
-10 - 7 = -17
-17 = -17 YUP!

x - y = 2
-5 - (-7) = 2
-5 +7 = 2
2 = 2 YUP!

b. you need to line your eqations up like you are going to add them together again. But when you like them up, it becomes clear that neither the x nor the y (like the y in part a) cancel out, so you need to mulitply one of the equations by a number that will allow either the x or y to cancel out. In this instance it is easiest to muliply the top equation by -2 so the 'y' in the first eqation becomes '-2y,' which cancels out with the +2y in the bottom eqation, leaving you with just the x to solve for!

3x + y = -15             -2 (3x + y = -15) <- make sure you multiply the entire thing by -2
+  x  + 2y = -10       +       x + 2y  = -10

then you have:
-6x - 2y = 30
+   x + 2y = -10
-5x        = 20
-5             -5
x = -4

step 2: now plug x into one of the equations to solve for y

x + 2y = -10
-4 + 2y = -10
+4           +4
2y = -6
2      2
y = -3

3x + y = -15
3(-4) + (-3) = -15
-12 + (-3) = -15
-12 - 3 = -15
-15 = -15 YUP!

x + 2y = -10
(-4) + 2 (-3) = -10
-4 + (-6) = -10
-4 - 6 = -10
-10 = -10 YUP!