MATH 135
SYSTEMS OF LINEAR EQUATIONS
1. Solve the following systems of linear equations by substitution and graph the
lines of the system.


2x + 3y = 6
2x − y = −1
(i)
(ii)
1
1
3
x − y
x + y =
=
2
2
2
i. Solve the second equation for x:
x = 12 + y.
Substitute that into the first equation:
2( 12 + y) + 3y = 6
1 + 2y + 3y = 6
1 + 5y = 6
5y = 5
y=1
Put y = 1 into the second equation:
x − 1 = 12
x = 32
The solution is ( 32 , 1).
ii. Solve the second equation for x:
x = 23 − 12 y
Substitute into first equation:
2( 32 − 21 y) − y = −1
3 − y − y = −1
−2y = −4
y=2
Put y = 2 into the second equation:
x + 1 = 23
x = 12 .
The solution is ( 12 , 2).
5
10
5
−4
−2
2
4
−4
−2
2
4
−5
−5
2x + 3y = 6
x − y = 21
2x − y = −1
x + 12 y = 32
2. Solve the following systems of linear equations by elimination and graph the
lines of the system.
(
(
x−y
=5
x+y =8
(ii)
(i)
−3x + 3y = 2
x−y =4
i. Add the two equations together to
eliminate y:
ii. Multiply the first equation by 3 and
add.
x+y =8
3x − 3y = 15
+ (x − y = 4)
+ (−3x + 3y = 2)
= 2x = 12
= 0 = 17
x=6
Put x = 6 into either equation:
6+y =8
y=2
The solution is (6, 2).
This equation is false regardless of
what x is, so there is no solution.
5
10
−4
−10
−5
5
10
−2
2
4
−5
−10
x+y =8
x−y =4
−10
x−y =5
−3x + 3y = 2
3. You have a data set of pairs (x, y) that includes the points (−1, −4), (1, 6),
and (3, 0). Furthermore, you know your data set is quadratic (it rises, reaches
one maximum value, and then falls). Find a reasonable quadratic model y =
ax2 + bx + c for your data.
We have three data points that supposedly fit on this quadratic. For each point,
take the quadratic model and make x equal to the first coordinate and y equal
to the second.
The point (−1, −4) gives a − b + c = −4
The point (1, 6) gives a + b + c = 6
The point (3, 0) gives 9a + 3b + c = 0
This is a system of three equations in three variables. Let’s try and solve it. I’d
subtract the second equation from the first:
a+b+c=6
− (a − b + c = −4)
= 2b = 10
b=5
Now that we know b = 5, let’s put that in the third equation:
9a + 15 + c = 0 ⇔ c = −9a − 15
Put c = −9a − 15 and b = 5 into the first equation:
a − 5 − 9a − 15 = −4 ⇔ −8a = 16 ⇔ a = −2
Finally,
c = 18 − 3 = 15
Therefore, the quadratic function that contains all three of the given data points
is
y = −2x2 + 5x + 3.
5
−2
−1
1
−5
−10
−15
2
3
4