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 \(y\).

\(3x+2y = 17\)

\(x – y = 4\)

Solve \(x\) and \(y\).

\(7x – 2y = 20\)

\(2x+y = 1\)

\(3x+2y = 17\)

\(x – y = 4\)

Solve \(x\) and \(y\).

\(7x – 2y = 20\)

\(2x+y = 1\)

**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.

\(4x+y = 14\)

\(3x+y = 3\)

Solve \(x\) and \(y\) using the combination method.

\(6x-2y = 9\)

\(7x+4y = 20\)

\(4x+y = 14\)

\(3x+y = 3\)

Solve \(x\) and \(y\) using the combination method.

\(6x-2y = 9\)

\(7x+4y = 20\)