Readiness Check for Algebra 1: Principles of Secondary Mathematics Solutions

1. $3\frac{1}{2}÷4\frac{1}{3}$

$\frac{7}{2}÷\frac{13}{3}=\frac{7}{2}·\frac{3}{13}=\frac{21}{26}$

2. $\frac{7}{8}–\frac{5}{12} \mathrm{LCD}\left(8, 12\right)=24$ 

$\frac{21}{24}–\frac{10}{24}=\frac{11}{24}$

3. $\left|6–10\right|+2^3–{\left(4+2\right)}^2÷6+\sqrt{81}$

$$\begin{gathered} \left|–4\right|+8–{\left(6\right)}^2÷6+9 \\ 4+8–36÷6+9 \\ 12–6+9 \\ 15 \end{gathered}$$

4. Evaluate: $–7^2$
   Is the answer positive or negative? Explain your reasoning.

–49
The answer is negative because $(–1)\left(7\right)\left(7\right)$ only has one negative number.

5. Evaluate: ${\left(–3\right)}^4$
   Is the answer positive or negative? Explain your reasoning.

81
The answer is positive because an even number of negatives multiplied together results in a positive value. $(–3)(–3)(–3)(–3) = 81$

6. At Floyd’s Family Farm the ratio of chickens to goats was seven to eight. If the farm owned 14 goats, how many chickens do they own? Show your work.

$$\begin{gathered} \frac{c}{g}=\frac{7}{8} \\ \frac{7}{8}=\frac{c}{14} \\ 8c=7\left(14\right) \\ \frac{8c}{8}=\frac{98}{8} \\ c=12\frac{2}{8}=12\frac{1}{4} \end{gathered}$$

The Farm owns 12 chickens, since you cannot have a fraction of a chicken.

7. Given $3x–4=17 \mathrm{and} \frac{1}{2}y+2=5$, find the sum of x and y. Show your work.

$\begin{array}{ccccc}3x–4=17& & \frac{1}{2}y+2= 5& & \mathrm{Sum} \mathrm{of} x \mathrm{and} y\\ +4 +4& & –2 –2& & x+y\\ \left(\frac{1}{3}\right)3x=21\left(\frac{1}{3}\right)& & \left(\frac{2}{1}\right)\frac{1}{2}y=3\left(\frac{2}{1}\right)& & 7+6\\ x=7& & y=6& & \overline{)13}\\ & & & & \end{array}$

8. Solve. Show all your work. Check your solution.
   $\begin{array}{rcl}& & \begin{array}{cc}–\frac{5}{2}\left(x–2\right)=15& \end{array}\end{array}$

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{ccc}–\frac{5}{2}x+5= 15 \\ –5 –5 \\ \left(\frac{–2}{5}\right)–\frac{5}{2}x=10\left(\frac{–2}{5}\right) \\ x=–4& \mathbf{\text{Check}} \\ –\frac{5}{2}\left(–4–2\right)=15 \\ –\frac{5}{2}\left(–6\right)=15 \\ 15=15& \end{array}\end{array} \end{gathered}$$

9. Find the error, describe it, and then solve the equation correctly.
   $\begin{array}{rcl}\frac{3}{4}x–5& =& 2\\ & & +5 +5\\ \frac{3}{4}x& =& 7\\ & & \overline{)–\frac{3}{4} –\frac{3}{4}}\\ x& =& 6\frac{1}{4}\\ & & \end{array}$

The highlighted step is incorrect. The reciprocal of $\frac{3}{4}$ should be multiplied on both sides instead of subtracting $–\frac{3}{4}$.

$x=7\left(\frac{4}{3}\right)=\frac{28}{3}$

10. Write an equation and solve.
    Three times a number (n) is equal to four times the same number minus two.

$$\begin{gathered} 3n= 4n–2 \\ –4n –4n \\ \left(–1\right)–1n=–2\left(–1\right) \\ n=2 \end{gathered}$$

11. Label the quadrants on the coordinate plane. Then, label the x– and y–axis.

12. $49$

$\overline{)1}, 7, \overline{)49}$

13. $48$

$\overline{)1}, 2, 3, \overline{)4}, 6, 8, 12, \overline{)16}, 24, 48$

14. Evalute $–2x^2{y}^3$ when $x=–3$ and $y=2.$

$$\begin{gathered} –2{\left(–3\right)}^2{\left(2\right)}^3 \\ –2\left(9\right)\left(8\right) \\ –144 \end{gathered}$$

15. Given $\begin{array}{rcl}3x–5& =& 10, \end{array}$what is $\begin{array}{rcl}& & 2–5x?\end{array}$

$\begin{array}{r}\begin{array}{ccc}3x–5=10& & 2–5x\\ +5 +5& & 2–5\left(5\right)\\ \left(\frac{1}{3}\right)3x=15\left(\frac{1}{3}\right)& & 2–25\\ x=5& & \overline{)–23}\end{array}\end{array}$

16. Using the values 1, 4, and 7 only once, find the combination that yields the smallest possible solution. Explain your thinking.

Solution: 

$\begin{array}{rcl}5& =& \frac{7}{1}x–4\\ & & +4 +4 \\ \left(\frac{1}{7}\right)9& =& 7x\left(\frac{1}{7}\right)\\ \frac{9}{7}& =& x\\ x& =& \frac{9}{7}\end{array}$

The smallest number comes from dividing by the largest possible number: 7.

17. Name the GCF and LCM of 12 and 15.

$$\begin{gathered} \begin{array}{ccc}\text{GCF}\left(12, 15\right)=3 \\ 12: 1, 2, \overline{)3}, 4, 6, 12 \\ 15: 1, \overline{)3}, 5, 15& & \text{LCM}\left(12, 15\right)=60 \\ 12, 24, 36, 48, \overline{)60}, ... \\ 15, 30, 45, \overline{)60}, ...\end{array} \end{gathered}$$

18. The length (l) of a rectangle is twice the width (w). Find the area when the perimeter of the rectangle is 54 units.

$\begin{array}{ccc}P=54, l=2w& & \\ P=2l+2w& & A=lw\\ 54=2\left(2w\right)+2w& & A=\left(18\right)\left(9\right)\\ 54=4w+2w& & A=162 {\mathrm{units}}^2\\ 54=6w& & \\ w=9 \mathrm{units}& & \\ l=2w=2\left(9\right)=18 \mathrm{units}& & \end{array}$

20. Solve. Show all your work. Checking your solution is required.

$\begin{array}{rcl}& & \begin{array}{ccc}3x+4–5x=1& & \\ –2x+4= 1& & \mathbf{\text{Check}}\\ –4 –4& & 3\left(\frac{3}{2}\right)+4–5\left(\frac{3}{2}\right)=1\\ \left(–\frac{1}{2}\right)–2x=–3\left(–\frac{1}{2}\right)& & \frac{9}{2}+\frac{8}{2}–\frac{15}{2}=\frac{2}{2}\\ x=\frac{3}{2}& & \frac{2}{2}=\frac{2}{2}\end{array}\end{array}$

22. If the volume of the pyramid is 32 and the length is 4 and the width is 2, find the height. Write the formula and solve. Use your formula sheet and show your work.

$\begin{array}{rcl}& & V=32, l=4, w=2\\ V& =& \frac{1}{3}lwh\\ 32& =& \frac{1}{3}\left(4\right)\left(2\right)h\\ \left(\frac{3}{8}\right)32& =& \frac{8}{3}h\left(\frac{3}{8}\right)\\ h& =& 12 \mathrm{units}\end{array}$

23. Complete the table of values for the equation: $y = 3x-4$

24. Describe the pattern for the x– and y–values from the table.

As the x–value increases by 1, the y–value increases by 3.