Alg 1 Part 2: Linear Combination HW

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NOTES
Notes: Solving Systems of Equations by Elimination
When solving a linear system, you can sometimes add or subtract the equations
to obtain a new equation in one variable. This method is called ELIMINATION.
Step 1:
Locate the variable with opposite or
same coefficients
Step 2:
same coefficients → Subtract
(multiply each term by -1 then add)
Solve the resulting equation
Step 3:
Substitute in either original
equation to find the value of the
other variable.
Step 4:
Name the solution
Step 5:
equations
Solve by linear combination (the elimination method)
1.
3x  y  8
7x  y  12
Example 1:
3x  4y   6
3x  2y  6
Example 2:
2x  y  7
2x  7 y  31
4.
3x  5y  7
4x  5y  14
5.
5y  7x  3
7x  y  9
7.
7x  2y  24
8x  2y  24
8.
5x  6y  10
5x  6y  46
9.
10.
9x  5y  23
12x  5y  19
11.
5x  6y  50
x  6y  26
12.
6.
1
y  8
2
3x  y  5
4x 
4
3
x  y 1
5
5
2x  3y   7
x  2y  7
x  2y  7
Sometimes the coefficients do not match up as well! You need to multiply first in order to be able to
eliminate one of the variables.
1.
x y 2
2x  7 y  9
2.
6x  2 y  1
2 x  3 y  5
4.
2x  5 y  3
4 x  10 y  6
7.
x y 0
1
1
x y 2
2
2
9.
18 x  6 y  24
3x  y  2
5.
9 y  5x  5
3x  7 y  5
6.
8.
10.
3
3
x  y  3
5
4
2
1
x y 8
5
3
9 x  15 y  24
6 x  10 y  16
4 x  8 y  10
x  2 y  6
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