Solving Systems of Equations Algebraically Solutions

Algebraically, systems of equations can be solved using substitution or elimination (linear combination) or a combination of both. 
Remember to solve for both variables and write them as an ordered pair.  
When using substitution : 
- First, replace the variable with a number or an expression. 
- Then, combine like terms to solve for the remaining variable.

When using elimination (or linear combinations): 
- First, decide which variable to eliminate from the equation. 
- Then, multiply one or both equations by the least common multiple and add the equations vertically. 
- Last, solve for the remaining variable. 
- Remember, the coefficients of the eliminated variable must have opposite values when adding them together (i.e., 5x and –5x).

Example 1

Solve the system of equations using substitution.

$a = \frac{1}{2} b - 1 4 a + 3 b = 15$

$\begin{array}{rcl} 4 (\frac{1}{2} b - 1) + 3 b & = & 15 \\ 2 b - 4 + 3 b & = & 15 \\ \frac{5 b}{5} & = & \frac{19}{5} \\ b & = & \frac{19}{5} \\ a & = & \frac{1}{2} (\frac{19}{5}) - 1 \\ a & = & \frac{19}{10} - \frac{10}{10} \\ a & = & \frac{9}{10} \end{array}$

$Solution : (\frac{9}{10}, \frac{19}{5})$

Check

$4 (\frac{9}{10}) + 3 (\frac{19}{5}) = 15$

$\begin{array}{rcl} \frac{36}{10} + \frac{57}{5} & = & 15 \\ \frac{18}{5} + \frac{57}{5} & = & 15 \\ \frac{75}{5} & = & 15 ✓ \end{array}$

Example 2

Solve the system using linear combinations.

$6 x + 2 y = 33.50 5 x + 7 y = 45.25$

$Eliminate x, LCM (5, 6) = 30 (5) (6 x + 2 y = 33.50) \Rightarrow 30 x + 10 y = 167.50 (- 6) (5 x + 7 y = 45.25) \Rightarrow - 30 x - 42 y = - 271.50 - 32 y = - 104 y = 3.25$

$\begin{array}{rcl} 6 x + 2 (3.25) & = & 33.50 \\ 6 x + 6.50 & = & 33.50 \\ 6 x & = & 27 \\ x & = & 4.50 \end{array}$

$(4.50, 3.25)$

Solving Systems of Equations Algebraically Solutions

Example 1

a=12b–14a+3b=15

Example 2

$a = \frac{1}{2} b - 1 4 a + 3 b = 15$