Practice Solutions

Solve.

Note

There is more than one way to solve a system of equations. If you have solved using another method, it should still result in the same solution.

1. $$\begin{gathered} 2x+5y=10 \\ x=3y+5 \end{gathered}$$
   
   $\begin{array}{rcl}2\left(3y+5\right)+5y& =& 10\\ 6y+10+5y& =& 10\\ 11y+10& =& 10\\ & & –10 –10\\ 11y& =& 0\\ y& =& 0\\ x& =& 3\left(0\right)+5\\ x& =& 5\\ & & \left(5, 0\right)\\ & & \end{array}$

2. $$\begin{gathered} x–y=14 \\ 4x+8y=116 \end{gathered}$$
   
   $$\begin{gathered} \left(–4\right)\left(x–y=14\right)\Rightarrow –4x+4y =–56 \\ \underline{+ 4x+8y= 116} \\ \frac{12y}{12}=\frac{60}{12} \\ y=5 \\ x–\left(5\right)=14 \\ x=19 \\ \left(19, 5\right) \end{gathered}$$

3. $$\begin{gathered} 6x+5y=–2 \\ 3x–2y=8 \end{gathered}$$
   
   $$\begin{gathered} 6x+5y= –2 \\ \left(–2\right)\left(3x–2y=8\right)\Rightarrow \underline{–6x+4y=–16} \\ \frac{9y}{9}=\frac{–18}{9} \\ y=–2 \end{gathered}$$
    $$\begin{gathered} 6x+5\left(–2\right)=–2 \\ 6x–10=–2 \\ +10 +10 \\ \frac{ 6x}{6}=\frac{8}{6} \\ x=\frac{4}{3} \\ \left(\frac{4}{3}, –2\right) \end{gathered}$$

4. $$\begin{gathered} y=2x–2 \\ 3x–y=6 \end{gathered}$$
   
    $\begin{array}{rcl}3x–\left(2x–2\right)& =& 6\\ 3x–2x+2& =& 6\\ x+2& =& 6\\ & & –2 –2\\ x& =& 4\\ y& =& 2\left(4\right)–2\\ y& =& 6\\ & & \left(4, 6\right)\end{array}$

Solve.

5. $$\begin{gathered} 3x+5y=15 \\ y=3x–6 \end{gathered}$$ 
   
   $\begin{array}{rcl}3x+5\left(3x–6\right)& =& 15\\ 3x+15x–30& =& 15\\ 18x–30& =& 15\\ & & +30 +30\\ \frac{18x}{18}& =& \frac{45}{18}\\ x& =& \frac{5}{2}=2.5\\ y& =& 3\left(2.5\right)–6\\ y& =& 1.5\\ & & \left(2.5, 1.5\right)\end{array}$

6. $$\begin{gathered} y=\frac{1}{2}x+7 \\ 5x+2y=2 \end{gathered}$$ 
   
   $\begin{array}{rcl}5x+2\left(\frac{1}{2}x+7\right)& =& 2\\ 5x+x+14& =& 2\\ 6x+14& =& 2\\ & & –14 –14\\ \frac{6x}{6}& =& \frac{–12}{6}\\ x& =& –2\\ y& =& \frac{1}{2}\left(–2\right)+7\\ y& =& 6\\ & & \left(–2, 6\right)\end{array}$

7. $$\begin{gathered} x–7y=8 \\ 4x+2y=–10 \end{gathered}$$
   
   $$\begin{gathered} \left(–4\right)\left(x–7y=8\right)\Rightarrow –4x+28y=–32 \\ \underline{4x+ 2y=–10} \\ \frac{30y}{30}=\frac{–42}{30} \\ y=–\frac{7}{5} \end{gathered}$$
   $\begin{array}{rcl}4x+ 2\left(–\frac{7}{5}\right) & =& –10\\ 4x–\frac{14}{5}& =& –\frac{50}{5}\\ & & +\frac{14}{5} +\frac{14}{5}\\ \left(\frac{1}{4}\right)4x& =& –\frac{36}{5}\left(\frac{1}{4}\right)\\ x& =& –\frac{9}{5}\\ & & \left(–\frac{9}{5}, –\frac{7}{5}\right)\end{array}$

8. $$\begin{gathered} 5x+12y=18 \\ 2x–8y=4 \end{gathered}$$
   
   $$\begin{gathered} \left(5x+12y=18\right)\left(2\right)\Rightarrow 10x+24y= 36 \\ \left(2x–8y=4\right)\left(–5\right)\Rightarrow \underline{–10x+40y=–20} \\ \frac{64y}{64}=\frac{16}{64} \\ y=\frac{1}{4} \end{gathered}$$
   $\begin{array}{rcl}2x–8\left(\frac{1}{4}\right)& =& 4\\ 2x–2& =& 4\\ 2x& =& 6\\ x& =& 3\\ & & \left(3,\frac{ 1}{4}\right)\end{array}$