An equation with two unknown has two variables. For a linear equation with 2 unknowns, we can solve two variables if we are given two equations.

For example, if we are only given $$2x + y = 5$$, we CANNOT solve it! We need one more equation to solve for both $$x$$ and $$y$$.

If we have two equations, namely $$2x + y = 5$$ and $$6x-y = 3$$, we can solve for both $$x$$ and $$y$$!

If we plot the two equations on a graph, it shows as this:

Notice that the lines converge at (1, 3)! It means for $$2x + y = 5$$ and $$6x-y = 3$$, the solution is $$x=1$$ and $$y=3$$. (where the lines meet)

## Method 1 – Substitution

The basic rule:  Express one unknown in terms of the other!   What do we mean by this? See an example below.

Example Solve for $$x$$ and $$y$$.
Equation (1) $$\quad 3x-2y=6$$
Equation (2) $$\quad x-y=1$$

Solution  From Equation (2), express the equation in terms of $$x$$.
$$x-y=1 \\ \, \\ x-y {\color{red}{+y}}=1 {\color{red}{+y}} \\ \, \\ x=1+y \\ \, \\$$
Now substitute this $$x$$ value into Equation (1).
$$3x-2y=6 \\ \, \\ 3 \underbrace{(1+y)}_{\substack{\text{From} \\ \text{Equation (2)}}} – 2y = 6 \\ \, \\ \underbrace{3+3y-2y}_{\text{Expand}} = 6 \\ \, \\ 3+y = 6 \\ \, \\ 3+y {\color{red}{-3}} = 6 {\color{red}{-3}} \\ \, \\ y = 3 \\ \, \\$$
We are not done yet! We need to find the value of $$x$$.
From Equation (2), substitute the value of $$y$$ we just found.
$$x=1+y \\ \, \\ x = 1+ 3 \\ \, \\ x = 4 \\ \, \\$$
Answer: $$x=4$$, $$y=3$$

Solve $$x$$ and $$y$$.
$$3x+2y = 17$$
$$x – y = 4$$
Let
Equation (1) $$\quad 3x+2y = 17$$
Equation (2) $$\quad x – y = 4$$

From Equation (2), express the equation in terms of $$x$$.
$$x – y = 4 \\ \, \\ x-y {\color{red}{+y}}=4 {\color{red}{+y}} \\ \, \\ x=4+y \\ \, \\$$
Now substitute this $$x$$ value into Equation (1).
$$3x+2y = 17 \\ \, \\ 3 \underbrace{(4+y)}_{\substack{\text{From} \\ \text{Equation (2)}}} + 2y = 17 \\ \, \\ \underbrace{12 + 3y}_{\text{Expand}} +2y = 17 \\ \, \\ 12+5y = 17 \\ \, \\ 12+5y {\color{red}{-12}} = 17 {\color{red}{-12}} \\ \, \\ 5y = 5 \\ \, \\ \dfrac{5y}{{\color{red}{5}}} = \dfrac{5}{{\color{red}{5}}} \\ \, \\ y = 1 \\ \, \\$$
Substitute $$y$$ into either Equation (1) or Equation (2) to find $$x$$.
$$x=5, y=1$$

Solve $$x$$ and $$y$$.
$$7x – 2y = 20$$
$$2x+y = 1$$
Let
Equation (1) $$\quad 7x – 2y = 20$$
Equation (2) $$\quad 2x+y = 1$$

From Equation (2), express the equation in terms of $$y$$ because it’s easier.
$$2x+y = 1 \\ \, \\ 2x+y {\color{red}{-y}}=1 {\color{red}{-y}} \\ \, \\ y = 1 – 2x \\ \, \\$$
Now substitute this $$y$$ value into Equation (1).
$$7x – 2y = 20 \\ \, \\ 7x- 2 \underbrace{(1 – 2x)}_{\substack{\text{From} \\ \text{Equation (2)}}} = 20 \\ \, \\ 7x \underbrace{-2+ 4x}_{\text{Expand}} = 20 \\ \, \\ 11x – 2 = 20 \\ \, \\ 11x – 2 {\color{red}{+2}} = 20 {\color{red}{+2}} \\ \, \\11x = 22 \\ \, \\ \dfrac{11x}{{\color{red}{11}}} = \dfrac{22}{{\color{red}{11}}} \\ \, \\ x = 2 \\ \, \\$$
Substitute $$x$$ into either Equation (1) or Equation (2) to find $$y$$.
$$x=2, y=-3$$

## Method 2 – Combining Equations

You can also solve $$x$$ and $$y$$ with combination method. Add or subtract the equations to eliminate a variable!
Example Solve for $$x$$ and $$y$$.
Equation (1) $$\quad x-2y=7$$
Equation (2) $$\quad 2x+2y=5$$

Solution  Notice that if we add Equations (1) and (2), we can eliminate $$y$$.
\begin{align*} \quad & x-2y=7 & \\ +\quad & \underline{2x+2y=5} & \color{red}{\text{add!}}\\ \quad & 3x=12 & \end{align*}

$$\require{cancel} \dfrac{\cancel{{\color{red}{3}}}x}{\cancel{{\color{red}{3}}}} = \dfrac{\cancelto{4}{{\color{red}{12}}}}{\cancel{{\color{red}{3}}}} \\ \, \\ x=4$$

Now substitute this $$x$$ value into either Equation (1) or (2).
$$x-2y=7 \quad \text{Equation (1)} \\ \, \\ \underbrace{4}_{\substack{x=4}} -2y = 7 \\ \, \\ 4-2y\color{red}{-4} = 7 \color{red}{-4} \\ \, \\ -2y = 3 \\ \, \\ y = -\dfrac{3}{2}$$

Answer: $$x=4$$, $$y = -\dfrac{3}{2}$$

Solve $$x$$ and $$y$$ using the combination method.
$$4x+y = 14$$
$$3x+y = 3$$
Because both equations have $$+x$$, subtracting may work well.
\begin{align*} \quad & 4x+y = 14 & \\ -\quad & \underline{3x+y = 3} & \color{red}{\text{subtract!}}\\ \quad & x = 11 & \end{align*}

Now substitute this $$x$$ value into either Equation (1) or (2).
$$4x+y = 14 \qquad \text{Equation (1)} \\ \, \\ \underbrace{(4)(11)}_{\substack{x=11}} +y = 14 \\ \, \\ 44+y = 14 \\ \, \\ 44+y\color{red}{-44} = 14 \color{red}{-44} \\ \, \\ y = -30$$

Answer: $$x=11$$, $$y = -30$$

Solve $$x$$ and $$y$$ using the combination method.
$$6x-2y = 9$$
$$7x+4y = 20$$
In order for the addition of two equations to work, we need to multiply Equation (1) by 2 to produce the same $$4y$$.
\begin{align*} \color{red}{2 \times} \quad & (6x-2y = 9) & \\ \quad & \underline{7x+4y = 20} & \\ \quad & \qquad ? & \end{align*}
\begin{align*} \quad & 12x -4y = 18 & \\ \color{red}{+}\quad & \underline{7x+4y = 20} & \color{red}{\text{add!}}\\ \quad & 19x = 38 & \end{align*}
$$\require{cancel} \dfrac{\cancel{{\color{red}{19}}}x}{\cancel{{\color{red}{19}}}} = \dfrac{\cancelto{2}{{\color{red}{38}}}}{\cancel{{\color{red}{19}}}} \\ \, \\ x=2$$
Now substitute this $$x$$ value into either Equation (1) or (2).
$$6x-2y = 9 \qquad \text{Equation (1)} \\ \, \\ \underbrace{(6)(2)}_{\substack{x=11}} -2y = 9 \\ \, \\ 12 – 2y = 9 \\ \, \\ 12-2y\color{red}{-12} = 9 \color{red}{-12} \\ \, \\ -2y = -3 \\ \, \\ y = \dfrac{3}{2}$$
Answer: $$x=2$$, $$y = \dfrac{3}{2}$$