# Equations and Inequalities-Target A-Prove It-Practice Problems

Target 2A: Use Algebraic Proofs to justify the steps to solve a linear equation or inequality including one solution, no solution or infinitely many solutions.

1) Solve the equation for x.
$\ \ \ \ \ \ \ \ 2x+5=3x-7$

2) Solve the inequality for y.
$\ \ \ \ \ \ \ \ 3y+7-12y>5y-8$

3) Solve the equation for p.
$\ \ \ \ \ \ \ \ \text{-}(2p-5)=13-4p$

4) Solve the inequality for h.
$\ \ \ \ \ \ \ \ \text{-}3(2h+1)<4h-3$

5) Solve the equation for j.
$\ \ \ \ \ \ \ \ \text{-}\dfrac{7}{2}j+13=\dfrac{1}{2}(\text{-}j+2)$

6) Solve the equation for x.
$\ \ \ \ \ \ \ \ 2(x+\dfrac{7}{2})=(x+3)+(x+4)$

7) Solve the inequality for k.
$\ \ \ \ \ \ \ \ \dfrac{3}{2}k-5\leq45-\dfrac{7}{2}k$

8) Solve the equation for m.
$\ \ \ \ \ \ \ \ 7m-(5m-12)=20+2m+4$

Drag the red circles onto the blue circles and see if you can identify the correct justifications for each step.
$\textbf{9)}$

$\textbf{10)}$

Looking at the work below, justify each step.
\begin{align*} \textbf{11)} \ \ 3x+7y&=12\\ \\ 7y&=\text{-}3x+12\\ \\ y&=\text{-}\dfrac{3}{7}x+\dfrac{12}{7}\\ \end{align*}

\begin{align*} \textbf{12)} \ \ \text{-}5x-7y+12x&=10+2y\\ \ \\ 7x-7y&=10+2y\\ \ \\ 7x-9y&=10\\ \ \\ \text{-}9y&=10-7x\\ \ \\ y&=\dfrac{10-7x}{\text{-}9}\\ \end{align*}

\begin{align*} \textbf{13)} \ \ \dfrac{3x}{5} &= \dfrac{11y}{10}\\ \ \\ 3x&=\dfrac{11y}{2}\\ \ \\ x&=\dfrac{11}{6}y\\ \end{align*}

\begin{align*} \textbf{14)} \ \ \text{-}(5x+2)&=12y+x\\ \\ \text{-}5x-2&=12y+x\\ \ \\ \text{-}6x-2&=12y\\ \\ 12y&=\text{-}6x-2 \\ y&=\text{-}\dfrac{1}{2}x-\dfrac{1}{6} \end{align*}