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Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
Module 11 Assessment Readiness
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Exercise 1 Page 426

a We are given a graph representing a system of linear equations, and need to determine whether each of the given equations is part of the system. To investigate this, we will follow a three step process.
  1. Express the given equations in slope-intercept form.
  2. Write the equations of the lines given in the graph in slope-intercept form.
  3. Compare the equations.
The slope-intercept form of a line is y=mx+b, where m is the slope and b the y-intercept. To write the equation 3x+y=5 in slope-intercept form, we will isolate the y-variable. 3x+y=5 ⇔ y=- 3x+5

Now, we will write the equation of the line given in the graph in the slope-intercept form as well. To do so, we will find the slope and the y-intercept of each one.

We can see in the graph that the slope of the blue line is 4 and the y-intercept is - 2. y=4x-2 Let's now find the equation of the red line.

We can see in the graph that the slope of the red line is - 3 and the y-intercept is 5. y=- 3x+5 The given equation, when written in slope-intercept form, matches the second equation from the system shown in the graph. Therefore, it is part of the system.

b To write the equation 2x+3y=8 in slope-intercept form, we will isolate the y-variable.
2x+3y=8
▼
Write in slope-intercept form
SubEqn

LHS-2x=RHS-2x

3y=- 2x+8
DivEqn

.LHS /3.=.RHS /3.

y=- 2x+8/3
WriteSumFrac

Write as a sum of fractions

y=- 2x/3+8/3
MovePartNumRight

a* b/c=a/c* b

y=- 2/3x+8/3
MoveNegNumToFrac

Put minus sign in front of fraction

y=- 2/3x+8/3
Therefore, the slope-intercept form of the second equation is as follows. y=- 2/3x+8/3 Looking back to Part A, we can tell that this equation does not match either of the equations of the graphed lines. Therefore, it is not part of the system.
c To write the equation - 8x+2y=- 4 using the slope-intercept form, we need to isolate the y-variable.
- 8x+2y=- 4
▼
Write in slope-intercept form
AddEqn

LHS+8x=RHS+8x

2y=8x-4
DivEqn

.LHS /2.=.RHS /2.

y=8x-4/2
WriteSumFrac

Write as a sum of fractions

y=8x/2-4/2
MovePartNumRight

a* b/c=a/c* b

y=8/2x-4/2
CalcQuot

Calculate quotient

y=4x-2
Therefore, the slope-intercept form of the third equation is as follows. y=4x-2 Once more, referring back to Part A. We can see that this equation matches that of the first line graphed in the system. Therefore, it is part of the system.
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Exercises 2 (Page 426)
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