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# 5.3: Quiz I

Difficulty Level: At Grade Created by: CK-12

Multiple Choice – Please circle the letter of the correct answer and write that letter in the space provided to the left of each question.

1. __________ The point (1, 3) is the solution to which of the following system of equations?

(a) $\begin{Bmatrix}x + 3y = 7 \\-x + 4y = 7\end{Bmatrix}$

(b) $\begin{Bmatrix}x + y & = 9 \\2x - y = -3\end{Bmatrix}$

(c) $\begin{Bmatrix}5x - 7y = -16 \\2x + 8y = 26\end{Bmatrix}$

(d) $\begin{Bmatrix}5x - 3y = 24 \\3x + 5y = 28\end{Bmatrix}$

1. __________ What is the solution of the following system of equations?

$\begin{Bmatrix}y = \frac{1}{2}x + 3 \\y = \frac{1}{6}x + 9\end{Bmatrix}$

(a) (3, 9)

(b) (12, 18)

(c) (-3, -9)

(d) (18, 12)

1. __________ Which system of linear equations has the following solution?

(a) $\begin{Bmatrix}2x - 5y = 5 \\4x + 2y = 22\end{Bmatrix}$

(b) $\begin{Bmatrix}x - 4y = -16 \\-4x + 5y = 15\end{Bmatrix}$

(c) $\begin{Bmatrix}x + 3y = -11 \\3x + 2y = 30\end{Bmatrix}$

(d) $\begin{Bmatrix}2x + y - 7 = 0 \\x - 2y - 1 = 0\end{Bmatrix}$

1. __________ What is the $x-$value for the solution in the following system of equations:

$\begin{Bmatrix}y = 3x + 11 \\y = -2x - 4\end{Bmatrix}$

(a) $x = -15$

(b) $x = 15$

(c) $x = 3$

(d) $x = -3$

1. __________ What is a system of linear equations that has no solution called?
1. independent
2. consistent
3. inconsistent
4. dependent
2. __________ What equation would be used to write equivalent systems for the following system of linear equations?

$\begin{Bmatrix}3x-6y=-7 \\5x+9y=-18\end{Bmatrix}$

(a) $-2x-15y-11=0$

(b) $8x+3y+25=0$

(c) $8x-3y = 25$

(d) $8x + 3y = 25$

1. __________ What is the solution for the following system of equations? $y=\frac{5}{6}x-2$ and $y=\frac{2}{9}x+9$
1. $l_1 \cap l_2 @ (18,13)$
2. $l_1 \cap l_2 @ \left(1, - \frac{7}{6}\right)$
3. $l_1 \cap l_2 @ (1,13)$
4. $l_1 \cap l_2 @ \left(\frac{5}{6}, \frac{2}{9}\right)$
2. __________ How many solutions are there for a system of linear equations that is consistent and independent?
1. two
2. infinite number
3. one
4. none

Answer the following questions in the space provided. Show all work.

1. Solve by substitution:

$\begin{Bmatrix}2x+y=-1 \\3x-2y=9\end{Bmatrix}$

1. Solve by elimination:

$\begin{Bmatrix}\frac{3}{4}x+\frac{5}{4}y=4 \\\frac{1}{2}x+\frac{1}{3}y=\frac{5}{3}\end{Bmatrix}$

1. C
2. D
3. A
4. D
5. C
6. B
7. A
8. C

$\begin{Bmatrix}2x+y=-1 \\3x-2y=9\end{Bmatrix}$

Solution:

$& 2x+y=-1\\& 2x-2x+y=-1-2x\\& \boxed{y=-1-2x}\\\\& 3x-2y=9 && 2x+y=-1\\& 3x-2(-1-2x)=9 && 2(1)+y=-1\\& 3x+2+4x=9 && 2+y=-1\\& 7x+2=9 && 2-2+y=-1-2\\& 7x+2-2=9-2 && \boxed{y=-3}\\& 7x=7\\& \frac{7x}{7} =\frac{7}{7}\\& \frac{\cancel{7}x}{\cancel{7}}=\frac{7}{7}\\& \boxed{x=1} && \boxed{l_1 \cap l_2 @ (1,-3)}$

$\begin{Bmatrix}\frac{3}{4}x+\frac{5}{4}y=4 \\\frac{1}{2}x+\frac{1}{3}y=\frac{5}{3}\end{Bmatrix}$

Solution:

$& \frac{3}{4}x+\frac{5}{4}y=4\\& 4 \left(\frac{3}{4}x\right)+4 \left(\frac{5}{4}y\right)=4(4)\\& \cancel{4} \left(\frac{3}{\cancel{4}}x\right)+\cancel{4} \left(\frac{5}{\cancel{4}}y\right)=4(4)\\& \boxed{3x+5y=16}\\& \frac{1}{2}x+\frac{1}{3}y=\frac{5}{3}\\& 6 \left(\frac{1}{2}x\right)+6 \left(\frac{1}{3}y\right)=6 \left(\frac{5}{3}\right)\\& \overset{3}{\cancel{6}} \left(\frac{1}{\cancel{2}}x\right)+\overset{2}{\cancel{6}}\left(\frac{1}{\cancel{3}}y\right)=\overset{2}{\cancel{6}}\left(\frac{5}{\cancel{3}}\right)\\& \boxed{3x+2y=10}\\& 3x+5y=16\\& 3x+2y=10\\& -1(3x+5y=16) \rightarrow -{\cancel{3x}}-5y=-16\\& 3x+2y=10 \rightarrow \qquad \quad \ \underline{\cancel{3x}+2y=10 \;\;\;}\\& \qquad \qquad \qquad \qquad \qquad \qquad \frac{-3y}{-3}=\frac{-6}{-3}\\& \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ {\color{red}y=2}\\& 3x+2y=10\\& 3x+2(2)=10\\& 3x+4=10\\& 3x+4-4=10-4\\& 3x=6\\& \frac{3x}{3}=\frac{6}{3}\\& {\color{red}x=2} \qquad \qquad \qquad \qquad \boxed{l_1 \cap l_2 @ (2,2)}$

Jan 16, 2013

Jan 14, 2015