Download fourth and final quiz

1. Math 084 - Quiz 4 Solutions
(1) Below are two lines that form a linear system.
(a) List all solutions of the system.
A solution to a linear system looks like an intersection point of the lines. Looking at
the graph, the intersection point is (−2, 3).
(b) Write the system of linear equations that correspond to this system.
The first line contains the points (−2, 3) and (4, 4). So the slope of the line is
4−3
= 16 .
4 − (−2)
Now we can use this slope and the point (4, 4) to write the equation of the first line in
point-slope form:
y − 4 = 16 (x − 4)
y = 16 x +
⇒
10
3 .
The second line contains the points (−2, 3) and (3, 1). So the slope is
1−3
= − 25 .
3 − (−2)
Now we can use this and the point (3, 1) to write the equation:
y − 1 = − 52 (x − 3)
y = − 25 x +
⇒
11
5 .
So the system of equations is
y = 16 x +
10
3
y = − 25 x +
11
5 .
(MORE ON BACK)
1
2
(2) Solve(the linear systems below.
x + 2y = 8
(a)
x − 3y = 3
Since both equations in this system are in standard form, I’ll use elimination. First,
I’ll multiply the second equation by −1, which gives me
(
x + 2y
=8
−x + 3y = −3
Now the x’s are set up to cancel, so I’ll add the equations to get
5y = 5
⇒
y = 1.
Now I’ll put 1 in for y in the equation x + 2y = 8:
x + 2(1) = 8
⇒ x = 6.
So the solution to the system of equations is (6, 1).
(
y = 4x + 5
(b)
−8x + 2y = 2
Since the first equation is already solved for y, I’ll use substitution. Putting 4x + 5
into the second equation for y, I get
−8x + 2(4x + 5) = 2
−8x + 8x + 10 = 2
10 = 2.
There is no value of x that makes this last line true, so there are no solutions to the
system of equations.