# MATH 10 Solving Systems of Linear Equations by tiw14488

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```									MATH 10              Solving Systems of Linear Equations

What is a SYSTEM OF EQUATIONS ? It’s any 2 or more equations considered together.
To solve a system of linear equations, we will find the point where the graphs of the two linear equations
intersect.

A.   Solve by GRAPHING

1) Where do the graphs of y = 2x + 1 and y = −3x + 6 intersect?
Step 1: draw the graph of y = 2x + 1 by marking              Step 2: draw the graph of y = −3x + 6 by marking
2                                                            3
the y-intercept at (0,1) and making slope .                the y-intercept at (0,6) and making slope − .
1                                                            1

5 .0
5.0

-5.0                            5.0
-5.0                                5.0

You can see clearly that the lines cross at (1,3). This is the solution of the system.

2)   Solve the system             x + 3y = 8
2y + 3x + 4 = 0

Try putting each equation in slope-intercept form to make them easier to graph.
x + 3y = 8                           2y + 3x + 4 = 0
−x         −x                              − 3x − 4 −3x − 4
3y − x + 8                        2y           −3x − 4
= 3                                       =
3                                 2              2
1      8                                    3
y = − 3 x + 3 (still ugly!)           y       = − 2 x − 2 (easy to graph!)
If slope-intercept form is still ugly, try getting
one convenient ordered pair.
1          8   3
e.g. if x = 5, y = − 3(5) + 3 = 3 = 1

5.0
1
Then make slope − 3 from that point.

You can see that the solution of the system
is (−4,4).                                                    -5.0                    5.0
If the two lines don’t meet at a convenient ordered pair, solving by GRAPHING can be very inconvenient.
ADDITION (or Linear-Combination) is the name given to one algebraic method of solving systems.

1)   The sum of 2 numbers is 6. If the smaller number is subtracted from the larger one, the difference
is 10. Find the numbers.

Call the numbers x and y. The first sentence is translated as         x + y = 6.
Pick one of them to be the smaller number. If y is smaller,
then the second sentence is translated as                             x − y = 10

To understand the “addition”, watch this:
If           =                            So if x + y = 6
and          =                            and   x − y = 10
then    + = +                             then 2x + 0 = 6 + 10
2x = 16              x=8

Then, since x + y = 6, we have 8 + y = 6, so y = −2.

The graphs of the two equations would cross at (8,−2). The answer to the problem is
“the 2 numbers are 8 and −2”. (Note: check and you’ll see that these work in the 2nd equation too.)

2)    3x + 4y = 6                                    3)   2x − y − 6 = 0
−3x + 3y = −20                                       5y − 2x = 10
ADD: 0 + 7y = −14                                     ADD: 0 + 4y − 6 = 0 + 10
y = −2                                                  +6       +6
4y     = 16, so y = 4
Then 3x + 4(−2) = 6
3x − 8 = 6                                    Then 2x − 4 − 6 = 0
+8      +8                                      2x − 10 = 0
3x        14
3      = 3                                            + 10      + 10
2x = 10, so x = 5
14
−
Solution: ( 3 ,−2)                                   Solution: (5,4)

NOTE: When you ADDED, one variable was eliminated. What if this doesn’t happen?

4)   2x − 5y = −17           If you ADD, you get 8x + 2y = −2 and you still have 2 variables.
6x + 7y = 15            Instead, multiply both sides of the first equation by −3.

Then you have          −3(2x − 5y) = (−17)−3 --------> −6x + 15y = 51
and the second equation --------------->                6x + 7y = 15

Now ADD                                                     0 + 22y = 66 x’s are eliminated
y = 3
Then 2x − 5(3) = −17
2x − 15 = −17
+ 15    + 15
2x = −2, so x = −1                                     −
Solution: (−1,3)
5) Sometimes you need to multiply in both equations so that one variable is eliminated.

5x − 4y = 14 multiply both sides by 2:      2(5x − 4y) = (14)2 ---->           10x − 8y = 28
2x + 3y = −22 multiply both sides by −5:   −5(2x + 3y) = (−22)(−5) --->       −10x − 15y = 110

Now ADD ------------->                                                0 − 23y = 138
y = −6
Then 5x − 4(−6) = 14
5x + 24 = 14
− 24    −24
5x         = −10, so x = −2             Solution: (−2, −6)

- multiply one or both equations so that one variable is eliminated
i.e., so that the x’s add up to zero, or the y’s add up to zero
- ADD the equations on the left side, on the right side
- solve the equation for the variable
- plug the value back into one of the original equations (Note: it doesn’t matter which one.)
- write the solution as an ordered pair (x,y) or give the answer in “words”.

C. Solve by SUBSTITUTION

1)   7x + y = 13         1st equation has just one y, which you can easily isolate: y = 13 − 7x
2x − 3y = 30

Since we are looking for the y-value that is the same in both equations, we can SUBSTITUTE our
isolated y into the second equation, as shown:         2x − 3(13 − 7x) = 30

Then solve for x:                                    2x − 39 + 21x   = 30
23x − 39     = 30
+ 39       + 39
23x      = 69
x     = 3
Then 7(3) + y = 13
21 + y = 13
−21         −21
y = −8                                                −
Solution: (3,−8)

2)   3x − 8y = −4
x + 6y = 16    Now it’s easiest to isolate x , in the 2nd equation: x = 16 − 6y

SUBSTITUTE into the first equation:           3(16 − 6y) − 8y = −4
48 − 18y − 8y = −4
−48               −48
− 26y = −52 , so y = 2
Then 3x − 8(2) = −4
3x − 16 = −4
+ 16 + 16
3x      = 12 , so x = 4                       Solution: (4,2)
SUMMARY OF STEPS FOR “SUBSTITUTION”
- find one equation where it’s easy to isolate one variable (usually where you have just one x or one y)
- substitute into the other equation
- solve the equation for the variable
- plug the value back into one of the original equations (Note: it doesn’t matter which one.)
- write the solution as an ordered pair (x,y) or give the answer in “words”.

D. Solve by COMPARISON

1)   6x + 1 + 11y = 0        Look for something that is the SAME in both equations.
11y + 5
= −2x           Do you see they both have 11y ? Isolate 11y in both equations.
4

11y + 5
1st equation: 11y = −6x − 1             2nd equation: 4( 4 ) = (−2x)4
11y + 5 = −8x
11y = −8x − 5

COMPARE the 11y in the 1st equation, the 11y in the 2nd equation – they must be equal!
−6x − 1 = −8x − 5
+ 8x        + 8x                    In practice is would have been
2x − 1 = −5                        easier to simply substitute for
+1       +1                    “11y” in the second equation…
2x = −4, so x = −2              Think of the COMPARISON
method as the SUBSTITUTION
Then 6(−2) + 1 + 11y = 0                           method used twice!
−12 + 1 + 11y = 0
−11 + 11y = 0
+ 11           + 11
11y = 11 , so y = 1                                  −
Solution: (−2,1)

2)   The sum of 4 times a number and 7 times a second number is 14. If double the first number is
subtracted from 7 times the second number, the result is −10. Find the two numbers.

First sentence:         4x + 7y = 14
Second sentence:        7y − 2x = −10         Notice 7y in both equations.

Isolate 7y in 1st equation: 7y = 14 − 4x
Isolate 7y in 2nd equation: 7y = 2x − 10

Therefore it must be true that 14 − 4x = 2x − 10
+ 4x + 4x
14      = 6x − 10
+ 10          + 10
24      = 6x       so x = 4

Then 4(4) + 7y = 14
16 + 7y = 14
− 16      − 16
2                                                 2
7y = −2 so y = − 7              Answer: The numbers are 4 and −7.
SUMMARY OF STEPS FOR “COMPARISON”
- find something the same in both equations
- isolate for what is the same
- write the two equations equal to each other
- solve for the variable
- plug the value back into one of the original equations (Note: it doesn’t matter which one.)
- write the solution as an ordered pair (x,y) or give the answer in “words”.

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