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Exercises Systems of equations

Example 4 from Algebra and Trigonometry, 11.1 Systems of Linear Equations: Two Variables

Solving a System by the Addition Method

Solve the given system of equations by addition.

\begin{array}{l} x + 2 y = −1 \\ - x + y = 3 \end{array}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Both equations are already set equal to a constant. Notice that the coefficient of $x$ in the second equation, –1, is the opposite of the coefficient of $x$ in the first equation, 1. We can add the two equations to eliminate $x$ without needing to multiply by a constant.

$\frac{\begin{array}{l} x + 2 y = - 1 \\ - x + y = 3 \end{array}}{3y=2}$
Now that we have eliminated $x,$ we can solve the resulting equation for $y .$
$\begin{array}{l} 3 y = 2 \\ y = \frac{2}{3} \end{array}$
Then, we substitute this value for $y$ into one of the original equations and solve for $x .$
$\begin{array}{l} - x + y = 3 \\ - x + \frac{2}{3} = 3 \\ - x = 3 - \frac{2}{3} \\ - x = \frac{7}{3} \\ x = - \frac{7}{3} \end{array}$
The solution to this system is $(- \frac{7}{3}, \frac{2}{3}) .$

Check the solution in the first equation.

$\begin{array}{l} x + 2 y = −1 \\ (- \frac{7}{3}) + 2 (\frac{2}{3}) = \\ - \frac{7}{3} + \frac{4}{3} = \\ - \frac{3}{3} = \\ −1 = −1 & True \end{array}$

How did it go?

PreviousSolving a System of Equations in Two Variables by Substitution Next Using the Addition Method When Multiplication of One Equation Is Required

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.1, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.