HUSD High School Algebra 1 2nd Semester Study Guide

HUSD High School Algebra 1 

2 nd Semester Study Guide 

1 Example: 

Finding x and y intercepts. 

Given: 2y + 4x = −8 

Set y = 0 to find x-intercept: 

Set x = 0 to find y-intercept: 

2y + 4x = −8 

2(0) + 4x = −8 

4x = −8 

x = −2 

2y + 4x = −8 

2y + 4(0) = −8 

2y = −8 

y = −4 

1´ Practice: 

Find the x and y intercepts. 

1) 6y + 2x =12 

2) 3x − 9y = 9 

The x-intercept is (−2,0) and the y-intercept is 

(0,−4). 

6.0 

2 Example: 

Graphing. 

Graph the equation 2y + 4x = −8 

Method 1: 

Find x and y-intercepts. 

From Example 1 (−2,0) and (0,−4) 

Plot the points and draw the line. 

Method 2: 

Make a table of values and plot the points. 

2´ Practice: 

Graph the given equation. 

3) 2x − 5y = −10 

4) y = 3x + 6 

5) 4y + 8 = −16x 

6) y = −2 

6.0 

Page 1 of 16 MDC@ACOE (HUSD) 05/01/10



3 Example: 

Slope = rise change in the y − coordinate 

= 

run change in the x − coordinate 

Horizontal Line: 

Find the slope of 

The line through 

Vertical Line: 

Find the slope of 

the line through 

(2,3) and (−1,3) (−2,−1) and (−2,3) 

m = 

= 

change in y 

change in x 

3− (3) 

−1− (2) 

m = 

= 

change in y 

change in x 

3− (−1) 

−2 − (−2) 

3´ Practice: 

Write the equation of the line in standard 

form: 

7) Find the equation of the line passing 

through the point ( 1, 5 ) with a slope of 

m = 0. 

8) Find the equation of the line passing 

− 1,4 with undefined 

through the point ( ) 

slope (or no slope). 

= 0 −3 

= 4 0 

Slope is m = 0 Since division by 0 

is not defined, this 

Equation is y = 3 line has no slope or 

undefined slope. 

Equation is x = −2 

7.0 




4 Example: 

Write an equation in slope-intercept form of 

the line passing through the point (6,−3) with 

slope m = 2. 

Use the point-slope form of a line, m is the 

slope and (x 1 

, y 1 

) = (6,−3) 

y − y 1 

= m(x − x 1 

) 

y − (−3) = 2(x − 6) 

y + 3 = 2x −12 

y + 3− 3 = 2x −12 − 3 

y = 2x −15 

4´ Practice: 

Write the equation of the line in 

slope-intercept form. 

9) line passes through (0,−5) and (3,4) 

10) line passes through (−3,1) and (5,5) 

For constants a and b: 

y = b is a horizontal line with slope m = 0 

x = a is a vertical line with no slope or 

undefined slope. 

7.0 

5 Example: 

Finding parallel and perpendicular slopes. 

The slope of line l is −2 which also can be 

written m = −2 

A line parallel to line l has the same slope so 

m = m = − 

1 

2 

A line perpendicular to line l has the slope 

m 2 

such that m⋅ 

m2 = −1. 

m⋅ 

m 

2 

−2⋅ 

m = −1 

2 

−2⋅m 

−2 

m 

2 

2 

= −1 

= − 1 

−2 

1 

= 

2 

Perpendicular slopes are opposite reciprocals. 

8.0 

5´ Practice: 

Line k is given by the equation 2x + 6y =12. 

11) Find the slope of the line parallel to 

line k. 

12) Find the slope of the line perpendicular 

to line k. 

13) Write the equation of the line that passes 

through the point ( −2,5) and is 

perpendicular to line k. 




6 Examples: 

Solving Equations by Factoring. 

a) 

( v + 7) ( v − 3) = 0 

( v + 7) = 0 or ( v − 3) = 0 

v = −7 or v = 3 

b) 

2x 2 +10x = 0 

2x(x + 5) = 0 

2x = 0 or (x + 5) = 0 

x = 0 or x = −5 

6´ Practice: 

Solve the equation by factoring. 

14) ( x −1) ( x + 6) = 0 

15) 

1 

z 

3 

2 

+ 5z 

= 0 

2 

16) 12m 

+ 4m 

= 40 

c) 

18n 2 − 27n = −9 

18n 2 − 27n + 9 = 0 

9(2n 2 − 3n +1) = 0 

9(2n −1)(n −1) = 0 

(2n −1) = 0 or (n −1) = 0 

2n =1 

n = 1 2 

or n =1 

14.0 




7 Example: 

Solve the system x − y = −4 

by graphing. 

y = −2x 

− 2 

1 st Graph x − y = −4 by finding intercepts. 

a. Use a table to find the intercepts. 

7´ Practice: 

Solve the system of equations by graphing. 

Check your solution algebraically by 

substituting the ordered pair in each equation 

to make sure it gives you a true statement in 

each equation. 

x x − y = −4 y 

0 0 − y = −4 4 

−y = −4 

−1(−y) = −1(−4) 

y = 4 

– 4 x − 0 = −4 

0 

x = −4 

b. Plot the intercepts (0,4) and (−4,0). 

c. Graph the line. (see graph below) 

2 nd Graph y = −2x 

− 2 using slope-intercept 

form y = mx + b . 

17) 

18) 

x + y = −4 

y = −2x + 5 

x − y = 5 

2x + 3y = 0 

a. Identify the slope m = − 2 1 

b. Identify the y-intercept 0,−2 ( ) 

c. Plot ( 0,−2), go down two units and right 

one unit and plot a point. Graph the line. 

3 rd Locate the point of intersection of the 

lines. 

( ) is the solution to the system 

∴ −2,2 

x − y = −4 

y = −2x 

− 2 check algebraically! 9.0 




8 Example: 

Solve the system of equations by the 

Substitution Method. 

Solve the system: 3 x + y = 5 

2x 

− y = 10 

1 st Solve for one variable (y in equation 1) 

3x 

+ y = 5 

3x + y − 3x = 5− 

3x 

y = 5− 

3x 

or y = − 3x 

+ 5 

2 nd Substitute − 3x + 5 for y in equation 2 

2x − (−3x + 5) =10 

3 rd Solve for x. 

2x − (−3x + 5) =10 

2x + 3x − 5 =10 

5x − 5 =10 

5x − 5 + 5 =10 + 5 

5x =15 

x = 3 

4 th Substitute x = 3 into y = − 3x 

+ 5 to solve 

for y. 

y = −3(3) + 5 

y = −9 + 5 

y = −4 

∴ ( 3,−4 ) is the solution to the system 

3x 

+ y = 5 

2x 

− y = 10 

8´ Practice: 

Solve the system of equations by the 

Substitution Method. 

19) 

20) 

−2x + 4y = −16 

x + 6y = −16 

y = 6x − 4 

y = −2x + 28 

9.0 




9 Example: 

Solve the system by the Elimination Method. 

9´ Practice: 

Solve the system of equations by using the 

Elimination Method. 

Solve the system: 

2x − y =1 

3x + 2y =12 

1 st Rewrite the equations so they have 

opposite terms. Multiply equation 1 by 2 so 

that −2y and 2y are opposites. 

2( 2x 

− y) = 2( 1) 

3x 

+ 2y 

= 12 

4x 

− 2y 

= 2 

3x 

+ 2y 

= 12 

2 nd Add vertically: 7x 

+ 0y 

= 14 

3 rd Solve for x: 

7x 

= 14 

x = 2 

4 th Solve for y by choosing one of the 

equations and substituting x = 2: 

3(2) + 2y =12 

6 + 2y − 6 =12 − 6 

2y = 6 

y = 3 

∴ ( 2,3) is the solution to the system 

2x − y =1 

3x + 2y =12 

9.0 

21) 

22) 

3x + 4y = 46 

−2x − 2y = −28 

3x + 2y = −9 

−10x + 5y = −5 




10 Example: 

Solving a Coin Problem using a system of 

equations. 

Alyssa has 25 coins that total $3.55. If she 

has only dimes and quarters, how many dimes 

does she have 

Let d = number of dimes 

Let q = number of quarters 

1 st Write two equations: 

Equation for the total number of coins: 

d + q = 25 

10´ Practice: 

Solve the following problem using a system 

of equations. 

23) A vending machine takes only nickels 

and dimes. At the end of the day there were 

three times as many nickels as dimes and a 

total of $25.00 in the machine. How many 

of each coin were there in the machine 

Equation for the total amount of money: 

0.10d + 0.25q = 3.55 

Now, the system of equations is: 

d + q = 25 

0.10d + 0.25q = 3.55 

2 nd Solve the system by the elimination 

method: 

Multiply equation 1 by −10 and multiply 

equation 2 by 100 to clear the decimals. 

−10d −10q = −250 

10d + 25q = 355 

Adding vertically gives the equation 

15q =105 

q = 7 

(solve for q) 

Choosing d + q = 25 and substituting q = 7 

Results in the following: 

d + q = 25 

d + 7 = 25 

d + 7 − 7 = 25 − 7 

d =18 

3 rd Answer the question: Alyssa has 18 

dimes. 

9.0 




11 Examples: 

Simplifying Radical Expressions. 

a) 

28 

9 

11´ Practice: 

Simplify. 

24) 

16 

⋅ 44 ⋅ 11 

64 

b) 

c) 

d) 

= 28 9 

= 22 ⋅ 7 

3 2 

= 2 7 

3 

18m 2 n 3 p 6 

= 3 2 ⋅ 2m 2 n 2 n( p 3 

) 2 

= 3mnp 3 2n 

= 

8x 

3x 3 

8x 

3x 3 

= 22 ⋅ 2 

3x 2 ⋅ 3 3 

= 22 ⋅ 2 ⋅ 3 

3 2 x 2 

= 2 6 

3x 

2 12 − 27 + 36 

25) − 3 7 

26) 48p 11 q 4 

27) −11 8 + 7 18 + 2 50 

= 2 2 2 ⋅ 3 − 3 2 ⋅ 3 + 6 2 

= 2 ⋅ 2 3 − 3 3 + 6 

= 4 3 − 3 3 + 6 

= 3 + 6 

2.0 




12 Examples. 

Solving Quadratic Equations. 

Solve by finding square roots. 

a) 

5y 2 −16 = 0 

5y 2 =16 

Solve for 

2 

y 

12´ Practice. 

Solve by finding square roots. 

28) ( x − 2) 2 = 49 

29) 3x 2 − 25 = 0 

y 2 = 16 5 

y = ± 16 5 

y = ± 4 5 

y = ± 4 5 ⋅ 5 

5 

y = ± 4 5 

5 

Solve by factoring. 

b) 

3w 2 + 5w = 2 

3w 2 + 5w − 2 = 0 

( 3w −1) ( w + 2) = 0 

3w −1 = 0 or w + 2 = 0 

Solve for y 

Rationalize 

the 

denominator 

Rewrite the 

equation in 

standard 

form 

Factor 

Zero Product 

Property 

Solve each 

equation 

Solve by factoring. 

30) 3( x + 6) ( x −1) = 0 

31) 4x 2 +12x = 7 

w = 1 3 or w = −2 14.0 




13 Examples. 

Solving Quadratic Equations. 

Solve by Completing the Square. 

a) 

x 2 + 8x +12 = 0 

⎛ 

x 2 + 8x + 8 ⎞ 

⎜ ⎟ 

⎝ 2⎠ 

x 2 + 8x = −12 

2 

⎛ 

= −12 + 8 ⎞ 

⎜ ⎟ 

⎝ 2⎠ 

x 2 + 8x + 4 2 = −12 +16 

( x + 4) ( x + 4) = 4 

( x + 4) 2 = 4 

x + 4 = ± 4 

x = −4 ± 2 

x = −4 + 2 or x = −4 − 2 

x = −2 or x = −6 

Solve by using the quadratic formula. 

b) 

3x 2 − 4x − 2 = 0 

a = 3, b = −4, c = −2 

x = −b ± 

x = − −4 

x = 

b2 − 4ac 

2a 

( ) ± −4 

2( 3) 

4 ± 16 + 24 

6 

x = 4 ± 40 

6 

x = 4 ± 2 10 

6 

x = 4 6 ± 2 10 

6 

x = 2 3 ± 10 

3 

x = 2 ± 10 

3 

( )2 − 4 3 

( )( −2) 

2 

Isolate 

2 

x and x 

Add half the linear 

coefficient 

squared 

Factor 

Rewrite as a 

square 

Solve for x 

Simplify the 

radical 

Standard form 

Identify the values 

of a, b, and c 

Substitute the values 

of a, b, and c 

Simplify the 

radicand 

Simplify 

14.0, 19.0 


Solve the quadratic equation by completing 

the square. 

32) x 2 − 2x = 5 

33) x 2 + 6x −11 = 0 

Solve the quadratic equation using the 

quadratic formula. 

34) 4x 2 +12x = 7 

35) 6x 2 + 5x − 40 = 0 




14 Example. 

Graphing Quadratic Functions. 

Graph y = 2x 2 + 4x −1 

1 st Find the vertex 

Identify a, b, and c; find the x-coordinate of 

the vertex. 

x = − b 

2a 

x = − 4 

2 2 ( ) 

x = −1 

2 nd Make a table of values. Choose x-values 

on both sides x = −1 

x y = 2x 

2 + 4x 

−1 

y 

–3 y = 2 −3 

–2 y = 2 −2 

–1 y = 2 −1 

0 y = 2 0 

1 y = 2 1 

( ) 2 + 4( −3) −1 5 

( ) 2 + 4( −2) −1 −1 

( ) 2 + 4( −1) −1 –3 

( ) 2 + 4( 0) −1 −1 

( ) 2 + 4( 1) −1 5 


Determine if the parabola is concave up or 

concave down. Justify your answer. 

36) y = 3x 2 −18 

37) y = −2x − x 2 +1 

Find the vertex of the parabola. 

38) y = x 2 − x − 2 

Graph the quadratic function. 

39) y = x 2 − x − 2 

40) y = −x 2 + 2x 

3 rd Graph the parabola 

21.0 




15 Examples. 

Simplifying Rational Expressions. 

a) 

2x 2 

3x ÷ 2x 2 

4 9x 7 

= 2x 2 

3x ⋅ 9x 7 

4 2x 2 

= 2⋅ 9⋅ x 2 ⋅ x 7 

3⋅ 2⋅ x 4 ⋅ x 2 

9 

2 ⋅ 3⋅ 3⋅ x 

= 

2⋅ 3⋅ x 6 

= 3x 3 

1 

= 3x 3 

b) 

m 2 + 5m + 6 

m 2 − 4 

= m + 2 

m + 2 

⋅ 

( m − 2) 2 

3m −12 

( ) 2 

3( m − 4) 

( )( m + 3) 

( )( m − 2) ⋅ m − 2 

= ( m + 2) 

( m − 2)2 

m + 3 

3 m + 2 

( ) 

( )( m − 2) ( m − 4) 

( )( m + 3) 

3( m − 4) 

= m − 2 

c) 

3d 

4d − 20 + d 6 

= 

= 

= 

3d 

2 ⋅ 2( d − 5) + d 

2 ⋅ 3 

2⋅ 3 ⋅ 2( d − 5) 

2( d − 5) 

( ) + 2⋅ d( d − 5) 

2 ⋅ 2 ⋅ 3( d − 5) 

3d 

2 ⋅ 2( d − 5) ⋅ 3 3 + d 

9d 

2 ⋅ 2 ⋅ 3 d − 5 

= 9d + 2d 2 −10d 

2 ⋅ 2 ⋅ 3 d − 5 ( ) 

= 

2d 2 − d 

2 ⋅ 2 ⋅ 3 d − 5 

( ) 

( ) 

( ) 

d 2d −1 

= 

2 ⋅ 2 ⋅ 3 d − 5 

or 

Multiply by the 

reciprocal 

Rewrite as one fraction 

Factor 

Find equivalent forms of 

one 

2d 2 − d 

12 d − 5 ( ) 

Factor 

Rewrite as one 

fraction 

Find equivalent 

forms of one 


Simplify. 

41) 

42) 

43) 

44) 

45) 

x 2 + 9x + 20 

x 2 −16 

( x + 4) x + 9 

x − 6 

( ) 

( ) 

7w − 5 

w − 2 − 2w + 5 

w − 2 

2x 

3 

5x 

5 + 2x 

+ 

20x 

÷ ( x + 4)2 

x − 6 

5 

v 2 + v − 6 − 6 

v 2 + 2v − 8 

12.0,13.0 




16 Example. 

Solving Rational Equations. 

Solve: 

4 

y − 2 − 2y − 3 

y 2 − 4 = 5 

y + 2 

4 

y − 2 − 2y − 3 

y − 2 

( )( y + 2) = 5 

y + 2 

( ) − ( 2y − 3) = 5( y − 2) 

4 y + 2 

4y + 8 − 2y + 3 = 5y −10 

2y +11 = 5y −10 

11 = 3y −10 

21 = 3y 

7 = y 

Note: Clear the denominators by multiplying 

by the LCD = (y − 2)(y + 2). 

Don’t forget to check you answer(s)! 


Solve the rational equation. 

46) 

47) 

5 

x + 2 = x 4 

1 

y 2 −16 − 2 

y + 4 = 2 

y − 4 

13.0 

17 Example. 

Linear Inequalities. 

a) Solve 1 < x − 4 < 9 

1 < x − 4 

5 < x 

and x − 4 < 9 

x 8 

3x > 6 

x > 2 

5.0 


Solve the inequality. 

48) −7 ≤ 3w − 4 ≤11 

49) 

1 

x + 3 < −9 or 4 − 4x ≤12 

2 




18 Examples. 

Absolute Value Equations and Inequalities. 

Solve: 

a) 

4 x −10 = 2 

4x −10 = 2 

b) 

4x =12 

x = 3 

x +1 < 2 

x +1< 2 

x −3 


Solve the absolute value equation. 

50) 3w − 4 + 3 =12 

Solve the absolute value inequality and graph 

your solution on a number line. 

51) 

1 

3 y + 4 >1 

52) 2x −1 ≤ 7 

-3 -2 -1 0 1 2 3 4 5 6 

c) 

2x −1 + 2 ≥ 5 

2x −1 ≥ 3 

2x −1≥ 3 

2x ≥ 4 

x ≥ 2 

or 

x ≤ −1 or x ≥ 2 

−( 2x −1) ≥ 3 

2x −1 ≤ −3 

2x ≤ −2 

x ≤ −1 

-3 -2 -1 0 1 2 3 4 5 6 

3.0 

End of Study Guide 




1) ( 0,2), ( 6,0) 

2) ( 0,−1), ( 3,0 ) 

3) 

4) 

5) 

24) 11 

Answers to Practice Problems: 

10) y = 1 2 x + 5 2 

11) m = − 1 3 

13) y − 5 = 3(x + 2) 

14) x = −6 or x =1 

12) m = 3 

15) z = −15 or z = 0 

16) m = −2 or m = 5 3 

17) lines intersect at ( 9,−13) 

18) lines intersect at ( 3,−2 ) 

19) ( 2,−3) same shape as 

y = x 2 

20) ( 4,20) 

21) ( 10,4 ) same shape as 

y = x 2 

22) ( −1,−3 ) 

23) 100 dimes, 300 nickels 

24) 11 25) − 21 

7 

33) x = −3 ± 2 5 

34) x = − 7 2 or x = 1 2 

35) x = 

−5 ± 985 

12 

36) concave up 

37) concave down 

38) vertex: 

⎛ 1 

2 ,− 9 ⎞ 

⎜ ⎟ 

⎝ 4⎠ 

⎛ 1 

39) vertex: 

2 ,− 9 ⎞ 

⎜ ⎟ , 

⎝ 4 ⎠ 

concave up 

40) vertex: ( 1,1 ), concave 

41) 

43) 5 

44) 

45) 

down 

x + 5 

x − 4 

42) 

2x 2 + 5x + 8 

20x 2 

−1 

(v + 3)(v + 4) 

46) x = −2 or x =10 

x + 9 

x + 4 

6) 

4 

26) 4 p 5 q 2 3p 

47) y = 1 4 

3 

2 

1 

-5 -4 -3 -2 -1 0 1 2 3 4 5 

-1 

-2 

-3 

-4 

7) y = 5 

8) x = −1 

9) y = 3x − 5 

27) 9 2 

28) x = −5 or x = 9 

29) x = ± 5 3 

3 

30) x = −6 or x =1 

31) x = − 7 2 or x = 1 2 

32) x =1 ± 6 

48) −1≤ w ≤ 5 

49) x < −24 or x ≥ −2 

50) w = − 5 3 or w = 13 3 

51) y < −15 or y > −9 

52) −3 ≤ x ≤ 4

HUSD High School Algebra 1 2nd Semester Study Guide

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