Seat # ____
Algebra 2
Name: ___________________________
Practice Problems – Chapter 3
Per: ______ Date: ____________
1)
On the coordinate plane below, PLOT the point ( –4, –6 )
Now draw a straight line through this point with Slope = 2
(Show how you determined which points to draw the line through)
Next, PLOT the point ( 4, –5 )
Draw a straight line through this point with Slope = –3
(Show how you determined which points to draw the line through)
Find the point of intersection for the two lines and LABEL IT CLEARLY.
2)
For each line you drew, find where the line intersects the Y-axis and use those intercepts (together with the
slopes) to determine the Slope-Intercept equation for each line. Write the Slope-Intercept form equations
below.
1
3)
Let’s do Problem #1 again. However, this time … let’s use a system of equations to find that point of
intersection. Do the specified work in the following order:
a) Convert your two linear equations from Slope-Intercept (as you wrote them in Problem #2) into
Standard Form (AX + BY = C) and write the two reformatted equations below.
b) Now use the space below to solve for X and Y in your two equations using either substitution or
elimination methods. You already know the answer and so you don’t have to check your answer.
But show all your work!
4)
4)
Without graphing, classify each system as independent, dependent, or
Inconsistent. Show work or explain HOW you reached your conclusion.
If independent then give the solution.
x + y = 3
 x + 3y = 9
2 x + y = 3
a) 
b) 
c) 
 y = 2x – 3
9y + 3x = 27
 y = –2x – 1
For the following system of equations, what solution would you get if you solved for X and Y? Don’t
bother working out the problem … just explain your answer and WHY this is your answer!
3Y – 2X = 24
3Y – 2X = 6
2
5)
Solve the follow systems of equations by graphing the equations and using the point of intersection to
determine the solution (the X and Y coordinates). Substitute your answers (the X and Y values) back into
BOTH equations to check.
a)
3X – 2Y = 12
X + 2Y = –4
b)
4X – 3Y = 12
3X – 6Y = –6
3
c)
2Y – 3X = 14
2X + 3Y = –5
d)
4X – 3Y = 1
3X – 4Y = –8
6)
Use either the substitution method or the elimination method to solve the following systems of equations.
Remember to check you answers by substituting the X and Y values back into BOTH of the original
equations!
a)
2X + 2Y = 20
5X – 2Y = 8
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CONTINUED: Use substitution or elimination to solve; check answers!
7)
b)
4X + 3Y = 17
4Y – 5X = 2
c)
2X + 4Y = –2
3Y + 3X = 3
Use the elimination method to solve the following systems of equations.
+ 4Y = 24
a)
b)
+
=
3X –
–
= 15
= 3
5
8)
Solve the following system of inequalities. Follow the instructions given below.
Eq. #1
Y≤ X–6
Eq. #2
Y
3 – 2X
Graph the two inequalities.
Lightly shade each inequality and then find
the overlapping shaded region.
Validate your answer by choosing points and substituting the (X,Y) values of the points back into the two
inequalities. Do this for a point …

outside both shaded areas – This point should FAIL for both inequalities.

outside the shaded area for Eq. #1 but inside the shaded area for Eq. #2 – This point should FAIL for
Eq. #1 and PASS for Eq. #2

inside the shaded area for Eq. #1 but outside the shaded area for Eq. #2 – This point should PASS
for Eq. #1 and FAIL for Eq. #2

inside the double shaded area – This point should PASS for both inequalities.
6
9)
Graph the absolute value
function:
Y =
Recall that the graphed
equation presents a visual
representation of all
X-Y value pairs which
satisfy the equation.
Also recall that when you
solve single-variable
equations or inequalities
(such as those below),
you are calculating the
X value or values which
satisfy the equation or
inequality.
Use your graph to determine the solutions for the following. Explain your logic.
= 12
≤ 15
≥ 9
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10) In this problem, you are asked to solve the following equality.
– 5 =
2
+ 5
You will do this by graphing both expressions as separate equations and then comparing the graphs to
each other.
Graph the following
equations
–5
Y = 2
Y =
+ 5
Locate all points of intersection and label them with their coordinates. For each solution point, take the Xvalue and Y-value of the coordinates and substitute those values back into BOTH of the equations. If your
graphs are correctly drawn, the X-values of the intersection points will be the solutions for the original problem
(solve the equality).
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CH 3 Practice Test ANSWERS
1. (1 pt) (1,4)
2. (2 pts)
Slope: 2 with point (-4,-6)
EQ A:
Slope: -3 with point (4,-5) EQ B:
3. (2 pts)
(a): EQ A:
EQ B:
(b): (1,4) show work
4. (3 pts)
(1st 4) (a) INDEPENDENT with explanation. (b) DEPENDENT with explanation. (c)
INCONSISTENT with explanation.
4. (1 pt) (2nd 4) INCONSISTENT with explanation
5. (4 pts) (a) (2,-3)
(b) (6,4)
(c) (-4,1)
(d) (4,5)
6. (3 pts) (a) (4,6)
(b) (2,3)
(c) (3,-2)
7. (2 pts) (a) (6,5)
(b) (2,-1)
8. (3 pts) (3,-3) with common region below both lines shaded; pick a test point in the region.
9. (2 pts) Vertex: (-2,0) with y-intercept (0,3).
(3 pts)
a) What is X when Y is 12? (6,12) and (-10,12).
b) What are X values between Y=-15 and Y=15 (
)
c) What are x values when
and
? (
)
10. (4 pts) (-6,1) and (3,7) should show intercepts. SHOW WORK where you plug in: X=-6 into original
problem and get a solution of 1; X=3 gets you 7. CHECK SOLUTIONS into BOTH equations
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