# Answer Key for California State Standards: Algebra I

 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula.

a. Write an equation involving only numbers that shows that the point
lies on the graph of the equation 2y = 6x - 5.

b. A line has a slope of and passes through the point (5, 8).
What is the equation for the line?

 The equation of the line must be of the form y = mx + b. The slope is given. Therefore, . To find the y intercept b , substitute the coordinates of the point (5,8) for x and y in the equation . This gives: Therefore This result may also be obtained by using the point-slope formula for a nonvertical line: and substituting ,, and .

 8.0: Students understand the concepts of parallel lines and perpendicular lines and how those slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point.

a. A line is parallel to the line for the equation:
. What is the slope of the parallel line?

 may be rewritten as y = x - 18, which has slope 1. Any line parallel to this one must have the same slope, 1.

b. What is the slope of a line perpendicular to the line for the
equation 3y = 7 - 6x ?

 3y = 7 - 6x may be rewritten as . The slope m of any line perpendicular to this one must satisfy m(-2) = -1. Therefore

c. What is the equation of a line passing through the point ( 7, 4 ) and
perpendicular to the line having the equation 3x - 4y - 12 = 0?

 The equation 3x - 4y - 12 = 0 may be rewritten as . The slope of this line is . The slope m of any line perpendicular to this one must satisfy . Therefore . So the equation of any perpendicular line must be of the form . Since the graph of the line contains the point (7,4), it is also true that So the answer is . This answer may also be obtained by using the point-slope formula with , , and .

 9.0: Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets.

a. Solve for the numbers x and y from the equations 2x - y = 1 and 3x - 2y = -1

 2x - y = 1 3x - 2y = -1 Multiply the first equation by -2 and then add it to the second equation -4x + 2y = -2 3x - 2y = -1 -x + 0y = -3 x = 3 Solve for x 2 · 3 - y = 1 Substitute x = 3 into the first equation and solve for y 6 - y = 1 y = 5 x = 3, y = 5 Solution

There are other ways to solve this problem. One can use one of the
equations to solve for one of the variables in terms of the other and
substitute that expression into the other equation. For example,
from 2x - y = 1, solving for y gives y = 2x -1. Substituting this
expression for y in the second equation gives

3x - 2(2x - 1) = -1
3x - 4x + 2 = -1
-x + 2 = -1
-x = -3
x = 3

Now substituting x = 3 into either equation yields y = 5. So the solution
again is x = 3 and y = 5.

b. Graph the equations 2x - y = 1 and 3x - 2y = -1 and circle the portion
of the graph which corresponds to the solution to the above
problem on your graph.

c. Graph the solution to the linear inequalities
2x - y > 1 and 3x - 2y < -1

 10.0: Students add, subtract , multiply, and divide monomials and polynomials . Students solve multistep problems, including word problems, by using these techniques.

a. Simplify

b. Let P = 2x2 + 3x - 1 and Q = -3x2 + 4x - 1

1. Calculate P + Q and collect like terms.

2. Calculate P - Q and collect like terms.

c. Calculate the product ( x2 - 1 ) ( 2x2 - x - 3 ) and collect like terms.

d. The area of a rectangle is 16. The length of the rectangle is
and the width is . What is x?

 A = length times width
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