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Solving System of Linear Equations: (lesson 1 of 5)

## Substitution Method

The substitution method is most useful for systems of 2 equations in 2 unknowns. The main idea here is that we solve one of the equations for one of the unknowns, and then substitute the result into the other equation.

Substitution method can be applied in four steps

Step 1:

Solve one of the equations for either x = or y = .

Step 2:

Substitute the solution from step 1 into the other equation.

Step 3:

Solve this new equation.

Step 4:

Solve for the second variable.

Example 1: Solve the following system by substitution

\begin{aligned} 2x + 3y &= 5 \\ x + y &= 5 \end{aligned}

Solution:

Step 1: Solve one of the equations for either x = or y = . We will solve second equation for y.

\begin{aligned} x + y &= 5 \\ \color{blue}{y} &\color{blue}{=} \color{blue}{5 - x} \end{aligned}

Step 2: Substitute the solution from step 1 into the second equation.

\begin{aligned} 2x + 3\color{blue}{y} &= 5 \\ 2x + 3\color{blue}{( 5 - x )} &\color{blue}{=} \color{blue}{5} \end{aligned}

Step 3: Solve this new equation.

\begin{aligned} 2x + 3( 5 - x) &= 5 \\ 2x + 15 - 3x &= 5 \\ - x + 15 &= 5 \\ - x &= 5 - 15 \\ \color{red}{x} &\color{red}{=} \color{red}{10} \end{aligned}

Step 4: Solve for the second variable

\begin{aligned} y &= 5 - \color{red}{x} \\ y &= 5 - \color{red}{10} \\ y &= - 5 \end{aligned}

The solution is: (x, y) = (10, -5)

Note: It does not matter which equation we choose first and which second. Just choose the most convenient one first!

Example 2: Solve by substitution

\begin{aligned} 2x + 5y &= 12 \\ 4x - y &= 2 \end{aligned}

Solution:

Step 1: Solve one of the equations for either x = or y =. Since the coefficient of y in equation 2 is -1, it is easiest to solve for y in equation 2.

\begin{aligned} 4x - y &= 2 \\ - y &= 2 - 4x \\ \color{blue}{y} &\color{blue}{=} \color{blue}{4x - 2} \end{aligned}

Step 2: Substitute the solution from step 1 into the second equation.

\begin{aligned} 2x + 5\color{blue}{y} &= 12 \\ 2x + 5 \color{blue}{(4x - 2)} &= 12 \\ \end{aligned}

Step 3: Solve this new equation ( for x ).

\begin{aligned} 2x + 5\color{blue}{(4x - 2)} &= 12 \\ 2x + 2x + 20x - 10 &= 12 \\ 22x &= 22\\ \color{red}{x} &\color{red}{=} \color{red}{1} \end{aligned}

Step 4: Solve for the second variable

\begin{aligned} y &= 4 \color{red}{x} - 2 \\ y &= 4 \cdot \color{red}{x} - 2 \\ y &= 2 \end{aligned}

The solution is: $(x, y) = (1, 2)$

Exercise: Solve the following systems by substitution

Level 1

 $$\color{blue}{x + y = 4}\\\color{blue}{2x - 3y = 18}$$ $( x , y ) = ( 6 , 2 )$ $( x , y ) = ( -6 , 2 )$ $( x , y ) = ( 6 , -2 )$ $( x , y ) = ( -6 , -2 )$

Level 2

 $$\color{blue}{3x + 5y = -2}\\\color{blue}{2x - y = 3}$$ $( x , y ) = ( 1 , -1 )$ $( x , y ) = ( -1 , -1 )$ $( x , y ) = ( 1 , 1 )$ $( x , y ) = ( -1 , 1 )$