Chapter 7 Notetaking Guide.pdf

Graphing Linear Systems 

Goal 

Estimate the solution of a system of linear equations 

by graphing. 

VOCABULARY 

System of linear equations 

Solution of a linear system 

Point of intersection 

Example 1 

Find the Point of Intersection 

Use the graph at the right to estimate 

the solution of the linear system. Then 

check your solution algebraically. 

x + 2y = -4 Equation 1 

x - 3y = 1 Equation 2 

Solution 

The lines appear to intersect once at 

( , ). 

-- -- 

t--+-+--t--t=t--t--t--+-+--i 

!-+--t--+-+3f-----1f--t--+--+--1 

h« + y - 4 

I ,......................- 

..... .-i-"'" 

1 3 x 

--.--~ 

x 31Y 1 ................ 

I 3 t-----1"""'....... 

Check Substitute for x and for y in each equation. 

x + 2y = -4 x - 3y = 1 

+ 2( ) J: -4 - 3( ) J: 1 

-4 1 

Answer Because ( , ) is a solution of each equation, 

( ) is the solution of the system of linear equations. 

=I-,I'--...- j t-,--I 

I -...... 

142 Algebra 1. Concepts and Skills Notetaklng Guide . Chapter 7

SOLVING A LINEAR SYSTEM USING GRAPH·AND-CHECK 

Step 1 Write each equation in a form that is 

------ 

Step 2 Graph both equations in the 

---------- 

Step 3 Estimate the coordinates 9f the 

-------~- 

Step 4 Check whether the coordinates give a solution by 

them into each equation of the 

::-;-------,---- --- 

linear system. 

A line in slopeintercept 

form, 

y = mx + b, has a 

slope of m and a 

y-intercept of b. 

Example 2 Graph and Check a Linear System 

Use the graph-and-check method to solve the linear system. 

5x + 4y = -12 Equation 1 

3x - 4y = -20 Equation 2 

1. Write each equation in slope-intercept form. 

Equation 1 Equation 2 

5x + 4y = -12 3x - 4y = -20 

4y = - 12 -4y = - 20 

y= y= 

2. Graph both equations. 

3. Estimate from the graph that the 

point of intersection is ( __,_). 

4. Check whether (_,_) is a 

solution by substituting for 

x and for y in each of the 

original equations. 

Equation 1 Equation 2 

5x + 4y = -12 3x - 4y = -20 

5( ) + 4( ) J: -12 3( ) - 4( ) J: - 20 

-12 -20 

Answer Because (__ , _) is a solution of each equation in the 

linear system, ( __, _) is a solution of the linear system. 

! y 

5r- 

3 I- 

1 f- 

-5 -3 -1 x 

r 1 I- 

r 3 I- 

Lesson 7.1 . Algebra 1 Concepts and Skills Notetaklng Guide 143

o Checkpoint Use the graph-and-check method to solve the 

linear system. 

1.3x - 4y = 4 2.5x + 2y = 4 

9x + 2y = 12 

x + 2y = 8 

-r y I y 

-1-7 f-f---l--f-+-+-+-+---l 

~--+-1--15'f---l-+-+-+-+--+-- 

-I-II---I-t--+-+-+-f--I--l 

1-1-3 t--I- --t--+---t---t--t---i 

I , 3 I 5 

I. 

-1-1 -+t--+-+-+-+--+--+--1 

-I 3 5 7 x 

7 x 

-:-3f-t--+--+-+-+-+-....+---l 

3. y = -2x - 3 4. Y = 3x + 4 

7x - 3y = -6 

2x + 5y = 25 

I 

}' 

"""I 

1-7 ,.......... I- 

1 I -2 

5 t--I- 

y 

2 

-4 2 4 x 

2 

3-I- 

4 

- 1- ._ f- ~ 

I 

- 

1 

1 

6 

-7 -5 1 x 

-3 i-I 

144 Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7

• 

So ving Li ear Systems 

by Subs itution 

Goal 

Solve a linear system by substitution. 

SOLVING A LINEAR SYSTEM BY SUBSTITUTION 

Step 1 Solve one of the equations for one"of its 

---- 

Step 2 Substitute the expression from Step 1 into the other 

equation and solve for the 

------ 

I Step 3 Substitute the value from into the revised equation 

from and sOllve. 

Step 4 Check the solution in each of the 

equations. 

Example 1 

Substitution Method: Solve for y First 

When you use 

I the substitution 

method, you can 

check the solution 

by substituting it for 

x and for y in each 

of the original 

equations. You can 

also use a graph to 

check your solution. 

Solve the linear system. 4x + Y = -5 Equation 1 

1. Solve for y in Equation 1. 

4x + Y = -5 

y = 

3x - y = 5 Equation 2 

Ori'ginal Equation 1 

Revised Equation 1 

2. Substitute for y in Equation 2 and find the 

value of x. 

3x - y = 5 Write Equation 2. 

3x - ( ) = 

---- 

5 Substitute for y. 

x + = 5 Simplify. 

x= Subtract from each side. 

x = 

Divide each side by 

3. Substitute for x in the revised Equation 1 and find the value 

of y. 

y= = 

----- 

4. Check that (_, __) is a solution by sUbstituting _ for x and 

fory in each of the original equations. 

Lesson 7.2 . Algebra 1 Concepts and Skills Notetaklng Guide 145 

\

When using 

substitution, you will 

get the same 

solution whether you 

solve for y first or x 

first. You should 

begin by solving for 

the variable that is 

easier to isolate. 

Example 2 Substitution Method: Solve for x First 

Solve the linear system. 

2x - 5y = -13 Equation 1 

x + 3y = -1 

Solution 

Equation 2 

1. Solve for x in Equation 2. 

x + 3y = -1 

Original Equation 2 

x= Revised 'Equation 2 

2. Substitute for x in Equation 1 and find the 

---- 

value of y. 

---- 

Answer The solution is ( , ). 

--- 

2x - 5y = -13 Write Equation 1. 

2( ) - 5y = -13 Substitute for x. 

____ - 5y = -13 

Use the distributive property. 

= -13 Combine like terms. 

Add 

to each side. 

Divide each side by 

3. Substitute for y in the revised Equation 2 and find the value 

of x. 

x= Write revised Equation 2. 

x= Substitute for y. 

x= Simplify. 

4. Check that ( , ) is a solution by substit,uting for x 

and for y in eachof the original equations. 


o Checkpoint Name the variable that you would solve for first. 

Explain. 

1.x - 2y = 0 I 2.4x + 2y = 10 

x - 8y = -5 7x - y = 12 

Use substitution to solve the linear system. 

3. y = x - 1 4. Y = -5x + 3 

1, x - 5y = -15 3x + 2y = -8 

Lesson 7.2 . A'igebra 1; Concepts and Skills Notetaking Guide 147

• 

Solving Linear Systems 

by Linear ombinat·ons 

Goal 

Solve a system of linear equations by linear combinations. 

VOCABULARY 

Linear combinations 

SOLVING A LINEAR SYSTEM BY LINEAR COMBINAT,IONS 

Step 1 Arrange the equations with 

terms in columns. 

Step 2 Multiply, if necessary, the equations by numbers to obtain 

coefficients that are 

for one of the variables. 

Step 3 

the equations from Step 2. Combining like terms 

with opposite coefficents will 

one variable. 

Solve for the 

-------- 

Step 4 Substitute the obtained from Step 3 into _ 

and 

--------- ---------- 

. Step 5 Check the solution in each of the equations. 

--- 

148 A'igebra 1 Concepts and Skills Notetaking Guide . Chapter 7

Example 1 

Add the Equations 


7x + 2y = -6 Equation 1 

5x - 2y = 6 Equation 2 

Solution 

Add the equations to get an equation in one variable. 

7x + 2y = -6 Write Equation 1. 

5x - 2y = 6 Write Equation 2. 

Add equations. 

Solve for 

Substitute for in the first equation and solve for 

7( ) + 2y = -6 Substitute for 

Solve for 

Check that (_, __ ) is a solution by substituting for x and 

for y in each of the original equations. 

Answer The solution is (_, __). 

o Checkpoint Use linear combinations to solve the system of 

linear equations. Then check your solution. 

1.4x + Y = -4 2.4x + 3y = 10 

-4x + 2y = 16 12x- 3y = 6 

Lesson 7.3 . Algebra 1 Concepts and Skills Notetaking Guide .149

Linear Systems and 

Problem Solving 

Goal 

Use linear systems to solve real-Ufe problems. 

Example 1 

Choosing a Solution Method 

Health Food A health food store mixes granola and raisins to malke 

20 pounds of raisin granola. Granola costs $4 per pound and raisins 

cost $5 per pound. How many pounds of each should be included for 

the mixture to cost a total of $85 

Solution 

Verbal 

Model 

Pounds of 

granola 

+ 

Pounds of 

raisins 

Total 

pounds 

Price of 

granola 

Pounds of 

granola 

+ 

Price of 

raisins 

Pounds 

• 

of raisins 

= 

Total 

cost 

Labels Pounds of granola =_ (pounds) 

Pounds of raisins = 

(pounds) 

Total pounds = 

(pounds) 

Price of granola =_ (dollars per pound) 

Price of raisins = 

(dollars per pound) 

Total cost ~ 

(dollars) 

Algebraic 

Model 

- 

+ - 

Equation 1 

+ = Equation 2 

Because the coefficients of x and yare 1 in Equation 1, 

is most convenient. Solve Equation for and 

------ 

the result in Equation . Simplify to obtain y = 

----- 

Substitute for y in Equation 1 and solve for x. 

Answer The solution is pounds of raisins and pounds of 

granola. 

Lesson 7.4 . Algebra 1 Concepts and Skills Notetaking Guide 151

Example 2 

Multiply then Add 


3x - 5y = 15 Equation 1 

2x + 4y = -1 

Solution 

Equation 2 

You can get the coefficients of x to be opposites by multiplying the 

first equation by and the second equation by 

3x - 5y = 15 Multiply by x- y= 

2x + 4y = -1 Multiply by 

x- y= 

Add the equations and solve for 

Substitute 

-- 

for in the second equation and solve for 

2x + 4y = -1 Write Equation 2. 

2x +- 4( ) = -1 Substitute for 

2x - = -1 Simplify. 

Solve for - 

Answer The solution is ( ). 

-- 

o Checkpoint Use linear combinations to solve the system of 

linear equations. Then check your solution. 

3. x - 3y = 8 4.6x + 5y = 23 

9x - 2y = -32 

3x + 4y = 11 

150 Algebra 1 Concepts and Skills Notetaklng Guide . Chapter 7 

.._--_.~----

Linear Systems and 

Problem Solving 

Goal 

Use linear systems to solve real-life problems. 

Example 1 

Choosing a Solution Method 

Health Food A health food store mixes granola and raisins to make 

20 pounds of raisin granola. Granola costs $4 per pound and raisins 

cost $5 per pound. How many pounds of each should be included for 

the mixture to cost a total of $85 

Solution 

Verbal 

Model 

Pounds of 

granola 

+ 

Pounds of 

raisins 

Total 

pounds 

Price of 

granola 

• 

Pounds of 

granola 

+ 

Price of 

raisins 

Pounds 

• 

of raisins 

i 

' - 

Total 

cost 

Labels Pounds of granola =_ (pounds) 

Pounds of raisins = 

Total pounds = 

Price of granola =_ 

Price of raisins = 

Total cost ~ 

(pounds) 

(pounds) 



(dollars) 

Algebraic + Equation 1 

- - 

Model 

+ Equation 2 

Because the coefficients of x and yare 1 in Equation 1, 

is most convenient. Solve Equation for and 

------ 

the result in Equation . Simplify to obtain y = 

----- 

Substitute for y in Equation 1 and solve for x. 

Answer The solution is pounds of raisins and pounds of 

granola. 


WAYS TO SOLVE A SYSTEM OF LINEAR EQUATIONS 

Substitution requires that one of the variables be 

on 

one side of the equation. It is especially convenient when one of 

the variables has a coefficient of or 

Linear Combinations can be applied to any system, but it is 

especially convenient when a 

appears in different 

equations with 

that are 

----- 

Graphing can provide a useful method for 

a solution. 

o Checkpoint Choose a method to solve the linear system. 

Explain your choice, and then solve the system. 

1. In Example 1, suppose the health food store wants to make 

30 pounds of raisin granola that will cost a total of $125. How 

many pounds of granola and raisins do they need Use the 

prices given in Example 1. 

152 Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7 

--_.--_._-------~------

• Special Types of Linear Systems 

al 

Identify how many solutions a linear system has. 

NUMBER OF SOLUTIONS OF A LINEAR SYSTEM 

If the two sollutions have 

slopes, then the system has 

---- 

one solution. 

Lines intersect: 

solution. 

----- 

If the two solutions have the slope but _ 

y-intercepts, then the system has no solution. 

Lines are parallel: 

solution. 

If the two equa.tions have the slope and the _ 

y-intercepts, then the system has infinitely many solutions. 

Lines coincide: 

solutions. 

x 


Example 1 

A Linear System with No Solution 

Show that the linear system has no solution. 

-x + y = -3 Equation 1 

-x + Y = 2 Equation 2 

Solution 

Method 1: Graphing Rewrite each equation in slope-intercept form. 

Then graph the linear system. 

y = Revised Equation 1 

y = Revised Equation 2 

3 

y 

I 

I 

I I 

[ 

1 

-5 -3 -[ [ 3 x 

3 

I 

[ 

I 

I 

I 

1 5 

Answer Because the lines have the same slope but different 

y-intercepts, they are 

lines do not 

, so the system has 

----' ----- 

Method 2: Substitution Because Equation 2 can be rewritten as 

y = , you can substitute for y in 

Equation 1. 

-x + Y = -3 Write Equation 1. 

-x + = -3 Substitute for y. 

Combine like terms. 

Answer The variables are 

and you are left with a 

statement that is regardless of the values of x and y. This 

tells you that the system has 

----- 


Example 2 

A Linear System with Infinitely Many Solutions 

Show that the linear system has many solutions. 

3x + y = -1 

Equation 1 

- 6x -' 2y = 2 Equation 2 

Solution 

Method 1: Graphing Rewrite each equation 

in slope-intercept form. Then 

graph the linear system. 

y = 

y = 



Answer From these equations you can see 

that the equations represent the same line. 

point on the line is a solution. 

--- 

Method 2: Linear Combinations You can multiply Equation 1 by 

Ox + Oy = 0 Multiply Equation 1 by . 

- 6x - 2y = 2 Write Equation 2. 

o = 0 Add equations. statement 

Answer The var'iables are 




----------- 

I Y 

I 5 1-. - f- - 

, 

-4 -2 2 

I 1 

3 

1 

.:- 3 I- - 

I 

I 

.- 

x 

o Checkpoint Solve the linear system and tell how many 

solutions the system has. 

1. x - 2y = 3 2. -2x + 3y = 4 

-5x + 10y = -15 -4x + 6y = 10 


Example 2 I A Linear System with Infinitely Many Solutions 

Show that the linear system has many solutions. 

3x + y = -1 

Equation 1 

- 6x - 2y = 2 Equation 2 

Solution 

Method 1: Graphing Rewrite each equation 

in slope-intercept form. Then 

graph the linear system. 

y = 

y = 



Answer From these equations you can see 

that the equations represent the same line. 

point on the line is a solution. 

--- 

Method 2: Linear Combinations You can multiply Equation 1 by 

Ox + Oy = 0 Multiply Equation 1 by 

- 6x - 

o 

2y = 2 Write Equation 2. 

= 0 Add equations. statement 

Answer The variables are 




----------- 

I 

i 

I 

y 

5 

3 

1 

I 

-4 -2 2 x 

I 1 

.~ 3 

II 

I 

o Checkpoint Solve the linear system and tell how many 

solutions the system has. 

1.x - 2y = 3 2. -2x + 3y = 4 

-5x + 10y = -15 -4x + 6y = 10 

Lesson 7.5 . Algebra 1 Concepts and Skills Notetaking Guide 

155

• Systems of Linear Inequalities 

Goal Graph a system of linear i'nequalities. 

VOCABULARY 

System of linear inequalities 

Solution of a system of linear inequalities 

GRAPHING A SYSTEM OF LINEAR INEQUALITIES 

Step 1 

the boundary lines of each inequality. Use a 

line if the inequality is < or > and a line if 

-- 

the inequality is :::; or ·z. 

Step 2 

the appropriate half-plane for each inequality. 

Step 3 

the solution of the system of inequalities as the 

intersection of the half-planes from Step 2. 

156 Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7 

-------------------~--------_.. _-_..._.. _--

Example 1 

Graph a System of Two Linear Inequalities 

\ 

 

 

To check your 

graph, choose a 

point in the overlap 

of the half-planes. 

Then substitute the 

coordinates into 

each inequality. If 

each inequality is 

true, then the point 

is a solution. 

Graph the system of linear inequalities. 

y - x ::::: -1 Inequality 1 

x + 2y < 1 Inequality 2 

Solution 

Graph both inequalities in the same 

coordinate plane. The graph of the 

system is the overlap, or 

------ 

of the two half-planes. 

Example 2 

J 

I I 3 

i 

y r~;--l 

1 

1 

+1 ~ 

1 

i 

-3 -1 1 I 3 xl 

T 1 

Graph a System of Three Linear Inequalities 

Graph the system of linear inequalities. 

y::::: - 3 Inequality 1 

x < 2 Inequality 2 

y < x + 1 Inequality 3 

I 3 

I 

J 

I 

I 

I 

I I I 

Solution 

The graph of y ::::: -3 is the half-plane 

and the line 

,. 

y 

3 

I 

i 

I 

The graph of x < 2 is the half-plane to 

the of the line 

The graph of y < x + 1 is the half-plane 

the 

line 

--- ----- 

I 

-3 

I 

1 

-1 

i 1 

1_- t3 

1 

-i 

3 , 

i 

I 

X 

Finally, the graph of the system is the 

, or , of the 

:-:----:---:-: 

three half-planes. 


Example 3 

Write a System of Linear Inequalities 

Write a system of inequalities that 

defines the shaded region at the right. 

Solution 

The graph of one inequality is the 

. half-plane to the left of 

--- 

The graph of the other inequality is 

the half-plane to the right of _ 

The shaded region of the graph is the vertical band that lies 

---- 

the two vertical lines, and , but not 

the Hnes. 

I 

3 

1 

Y 

-il 

i 

r--- 

-3 -1 

II 

I 

-t- ~~ 

! 

Answer The system of linear inequalities below defines the 

shaded region. 

Inequality 1 

Inequality 2 

o Checkpoint Complete the following exercises. 

1. Graph the system of linear 

inequalities. 

y --x - 1 

2 

I 

y 

3 

1 

-3 

-1 

,-I 

1 3 x 

+3 

I 

I 

2. Write a system of linear inequalities 

that defines the shaded region. 

I I 

I 

. 

I 

I 

I 

I 

5 

3 

y 

-3 -1 

i 1 1 3 x 

I 

158 Algebra 1 Concepts and Skills Notetaklng Guide . Chapter 7

Words to Review 

Give an example of the vocabulary word. 

System of linear equations 

Solution of a linear system 

Point of intersection 

Linear combination 

System of linear inequalities 

Solution of a system of Unear 

inequalities 

Review your notes and Chapter 7 by using the Chapter Review on 

pages 431-434 of your textbook. 

Words to Review . Algebra 1 Concepts and Skills Notetaking Guide 159

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