1 Solve $$p$$ and $$q$$.
$$5p + q = 7$$
$$7p – q = 17$$
$$\blacktriangleright$$ Substitution Method
From Equation (1), express the equation in terms of $$q$$.
$$5p + q = 7 \\ \, \\ 5p+q {\color{red}{-5p}}=7 {\color{red}{-5p}} \\ \, \\ q= 7-5p \\ \, \\$$
Now substitute this $$q$$ value into Equation (2).
$$7p – q = 17 \\ \, \\ 7p -\underbrace{(7-5p)}_{\substack{\text{From} \\ \text{Equation (1)}}} = 17 \\ \, \\ 7p \underbrace{-(7-5p)}_{\text{Expand}} = 17 \\ \, \\ 7p-7+5p = 17 \\ \, \\ 12p – 7 = 17 \\ \, \\ 12p-7 {\color{red}{+7}} = 17 {\color{red}{+7}} \\ \, \\ 12p=24 \\ \, \\ p=2 \\ \, \\$$
Since $$5p + q = 7$$ and $$p=2$$,
$$5p + q = 7 \\ \, \\ 5\underbrace{(2)}_{p=2} + q = 7 \\ \, \\ 10 + q = 7 \\ \, \\ 10+q {\color{red}{-10}} = 7 {\color{red}{-10}} \\ \, \\ q = -3 \\ \, \\$$
Answer: $$p=2$$ and $$q = -3$$
$$\blacktriangleright$$ Combination Method
Notice that if we add Equations (1) and (2), we can eliminate $$q$$.
\begin{align*} \quad & 5p + q = 7 & \\ +\quad & \underline{7p – q = 17} & \color{red}{\text{add!}}\\ \quad & 12p = 24 & \end{align*}

$$\require{cancel} \dfrac{\cancel{{\color{red}{12}}}p}{\cancel{{\color{red}{12}}}} = \dfrac{\cancelto{2}{{\color{red}{24}}}}{\cancel{{\color{red}{12}}}} \\ \, \\ p=2$$

Now substitute this $$p$$ value into either Equation (1) or (2).
$$5p + q = 7 \qquad \text{Equation (1)} \\ \, \\ 5\underbrace{(2)}_{\substack{p=2}} +q = 7 \\ \, \\ 10 + q = 7 \\ \, \\ 10 + q\color{red}{-10} = 7 \color{red}{-10} \\ \, \\ q = -3 \\ \, \\$$
Answer: $$p=2$$ and $$q = -3$$

2Solve $$x$$ and $$y$$.
$$4x-y=2$$
$$x+y=3$$
$$\blacktriangleright$$ Substitution Method
From Equation (2), express the equation in terms of $$x$$.
$$x+y=3 \\ \, \\ x+y{\color{red}{-y}}=3 {\color{red}{-y}} \\ \, \\ x=3-y \\ \, \\$$
Now substitute this $$x$$ value into Equation (1).
$$4x-y=2 \\ \, \\ 4\underbrace{(3-y)}_{\substack{\text{From} \\ \text{Equation (2)}}} -y = 2 \\ \, \\ \underbrace{4(3-y)}_{\text{Expand}} -y = 2 \\ \, \\ 12-4y-y = 2 \\ \, \\ 12-5y = 2 \\ \, \\ 12-5y {\color{red}{-12}} = 2 {\color{red}{-12}} \\ \, \\ -5y=-10 \\ \, \\ \dfrac{-5y}{-5} = \dfrac{-10}{-5} \\ \, \\ y=2 \\ \, \\$$
Since $$x+y=3$$ and $$y=2$$,
$$x+y=3 \\ \, \\ x+\underbrace{(2)}_{y=2} = 3 \\ \, \\ x+2 = 3 \\ \, \\ x+2 {\color{red}{-2}} = 3 {\color{red}{-2}} \\ \, \\ x=1 \\ \, \\$$
Answer: $$x=1$$ and $$y=2$$
$$\blacktriangleright$$ Combination Method
Notice that if we add Equations (1) and (2), we can eliminate $$y$$.
\begin{align*} \quad & 4x-y=2 & \\ +\quad & \underline{x+y=3} & \color{red}{\text{add!}}\\ \quad & 5x = 5 & \end{align*} \\ \, \\ x=1

Now substitute this $$x$$ value into either Equation (1) or (2).
$$x+y=3 \qquad \text{Equation (2)} \\ \, \\ \underbrace{(1)}_{\substack{x=1}} +y = 3 \\ \, \\ 1+y = 3 \\ \, \\ 1+y\color{red}{-1} = 3 \color{red}{-1} \\ \, \\ y=2 \\ \, \\$$
Answer: $$x=1$$ and $$y=2$$

3 Solve $$w$$ and $$y$$.
$$9w-2y = 16$$
$$2w + y = 5$$
$$\blacktriangleright$$ Substitution Method
From Equation (2), express the equation in terms of $$y$$.
$$2w + y = 5 \\ \, \\ 2w + y {\color{red}{-2w}}=5 {\color{red}{-2w}} \\ \, \\ y= 5-2w \\ \, \\$$
Now substitute this $$y$$ value into Equation (1).
$$9w-2y = 16 \\ \, \\ 9w -2\underbrace{(5-2w)}_{\substack{\text{From} \\ \text{Equation (2)}}} = 16 \\ \, \\ 9w \underbrace{-2(5-2w)}_{\text{Expand}} = 16 \\ \, \\ 9w – 10 + 4w = 16 \\ \, \\ 13w -10 = 16 \\ \, \\ 13w -10 {\color{red}{+10}} = 16 {\color{red}{+10}} \\ \, \\ 13w=26 \\ \, \\ w=2 \\ \, \\$$
Since $$2w + y = 5$$ and $$w=2$$,
$$2w + y = 5 \\ \, \\ 2\underbrace{(2)}_{w=2} + y = 5 \\ \, \\ 4+y=5 \\ \, \\ 4+y {\color{red}{-4}} = 5 {\color{red}{-4}} \\ \, \\ y=1 \\ \, \\$$
Answer: $$w=2$$ and $$y=1$$
$$\blacktriangleright$$ Combination Method
In order for the addition of two equations to work, we need to multiply Equation (1) by 2 to produce the same $$2y$$.
\begin{align*} \quad & 9w-2y = 16 & \\ \color{red}{2 \times}\quad & \underline{(2w + y = 5)} & \\ \quad & \qquad ? & \end{align*}

\begin{align*} \quad & 9w-2y = 16 & \\ \color{red}{+}\quad & \underline{4w+2y=10} & \color{red}{\text{add!}}\\ \quad & 13w = 26 & \end{align*} \\ \, \\ w=2

Now substitute this $$w$$ value into either Equation (1) or (2).
$$2w + y = 5 \qquad \text{Equation (2)} \\ \, \\ 2\underbrace{(2)}_{\substack{w=2}} +y = 5 \\ \, \\ 4+ y = 5 \\ \, \\ 4+y\color{red}{-4} = 5 \color{red}{-4} \\ \, \\ y=1 \\ \, \\$$
Answer: $$w=2$$ and $$y=1$$