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Precalculus MATH F156X Summer 2026

Dr. Latrice Bowman

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Section GR-4 Systems of Equations

Worksheet Systems of Equations

For this outcome Specifically, for this outcome, students should be able to:
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Systems of Linear Equations
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A system of linear equations is a set of two or more linear equations that are solved simultaneously.The solution to a system of linear equations is the set of values that satisfy all equations in the system and are denoted by an ordered pair (x, y) for a two-variable system or an ordered triple (x, y, z) for a three-variable system.
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A system of linear equations is consistent if it has at least one solution. It is inconsistent if it has no solutions. It is dependent if it has infinitely many solutions.
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Substitution Method
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To solve a system of linear equations using substitution, solve one equation for one variable and substitute that expression into the other equation. Solve for the remaining variable and then back-substitute to find the value of the first variable. If the system is inconsistent, no solution exists. If the system is dependent, there are infinitely many solutions. If there are infinitely many solutions, the solutions are expressed in terms of a parameter but still written as an ordered pair.
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Elimination Method
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To solve a system of linear equations using elimination, add or subtract equations to eliminate one variable. This is done by multiplying one or both equations by constants so that the coefficients of one variable are opposites. Then add or subtract the equations to eliminate that variable. Solve for the remaining variable and back-substitute to find the value of the eliminated variable.
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Nonlinear Systems
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A nonlinear system of equations is a system in which at least one equation is not linear. These systems can be solved using substitution or elimination methods, but may require more advanced techniques such as factoring or using the quadratic formula.In nonlinear systems, the number of solutions can vary depending on the specific equations involved.
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Worksheet GR-4 Systems of Linear Equations

Solving Systems by Substitution
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Solving Systems by Elimination
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Solving Three Variable Systems
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Solving Nonlinear Systems
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Section GR-4 Rubric

Worksheet Worksheet

Before attempting the GR-4 outcome, you should review the following rubric. This shows the criteria you will be graded on and the expectations to earn each mastery level.
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Table 17. GR4 Rubric
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Criteria M P R N
Simplifying Algebraic Expressions All answers are fully simplified. Exact answers are given, not approximations. Radical and exponential expressions are used appropriately and are simplified in the correct manner No algebraic errors OR some expressions are not simplified. There are some algebraic errors and some expressions were not simplified. There are some algebraic errors and some expressions were not simplified.
Problem Structure and Mathematical Communication Problems have a clear beginning, middle, and end. Work progresses clearly from one step to the next. The work provided uses appropriate methods and notation and provides a clear solution. Problems have clear beginnings and ends. Work progresses from one step to the next. The work provided uses algebraic methods, correct notation, and provides a solution. Parts of the mathematical structure are missing, or steps in the progress of the solution are missing. There are errors in the mathematical notation. The work provided uses algebraic methods and provides a solution. There is not a clear structure to the solution, notation is used incorrectly, and it is unclear what methods are being used.
Solve a 2-Variable System Using Substitution Correctly used properties of equality to rewrite one of the equations. Correctly substituted into the other equation. Found the correct values and gave them as an ordered pair. Used properties of equality to rewrite one of the equations with minor errors OR substituted into the other equation with minor errors Or found the and gave them as an ordered pair. Did not use substitution to solve the system; errors in the values found; did not give solution as an ordered pair. No evidence of solving the 2-variable system.
Solve a 3-Variable System Picked a base equation to eliminate a variable in the system or used correct row-reduction; Reduced the system to two variables and solved; used back substitution to find the correct ordered triple; gave the answer using correct notation. Picked a base equation to eliminate a variable in the system OR correctly used substitution to remove a variable; Reduced the system to two variables and solved; used back substitution to find an ordered triple; minor errors in at most one step. Used algebraic methods to find the values; errors in multiple steps; OR incorrectly substituted to try to solve the system. No evidence of solving the 3-variable system.
Solve a Non-linear System Used substitution or elimination as appropriate to isolate one variable; found the correct values for the variable and substituted to get the other variable; gave correct ordered pairs. Used substitution or elimination as appropriate to isolate one variable; found the correct values for the variable and substituted to get the other variable; found some ordered pairs; minor errors in at most one step. Incorrectly used substitution or elimination when the other method is more appropriate; errors in finding the values of the variables, or errors in giving the solution. No evidence of solving the nonlinear system.
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Worksheet GR-4 Systems of Linear Equations

This outcome covers systems of linear equations. You should be able to solve systems of linear equations using various methods such as substitution and elimination. You should also be able to determine the number of solutions to a system of linear equations and interpret the solutions in context.
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For the quiz on this outcome you would be given 30 minutes to complete 2 problems. Here are samples of the types of problems you will encounter:
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1.

Solve the system using substitution: \(\begin{cases} 2x + y = 5 \\ 3x - 2y = 8 \end{cases}\)
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Hint.
Rewrite the first equation to express y in terms of x. Then substitute this expression into the second equation.
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Answer.
\(\left(\frac{18}{7}, -\frac{1}{7}\right)\)
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Solution.
First, solve the first equation for y: \(y = 5 - 2x\text{.}\) Then substitute this into the second equation: \(3x - 2(5 - 2x) = 8\text{.}\) Simplify: \(3x - 10 + 4x = 8\text{.}\) Combine like terms: \(7x = 18\text{.}\) Solve for x: \(x = \frac{18}{7}\text{.}\) Finally, substitute back to find y:
\begin{align*} y & = 5 - 2 \cdot \frac{18}{7}\\ &= 5 - \frac{36}{7}\\ & = \frac{35 - 36}{7}\\ & = -\frac{1}{7} \end{align*}
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2.

Solve the system using elimination:
\begin{gather*} \begin{cases} 4x-3y=12\\ 3x+2y=13 \end{cases} \end{gather*}
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Hint.
Multiply the first equation by 2 and the second equation by 3 to eliminate y.
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Answer.
\(\left(\frac{63}{17}, \frac{4}{17}\right)\)
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Solution.
Multiply the first equation by 2: \(8x - 6y = 24\text{.}\) Multiply the second equation by 3: \(9x + 6y = 39\text{.}\) Add the two equations: \(17x = 63\text{.}\) Solve for x: \(x = \frac{63}{17}\text{.}\) Substitute back into the first equation: \(4\left(\frac{63}{17}\right) - 3y = 12\text{.}\) Simplify: \(\frac{252}{17} - 3y = 12\text{.}\) Solve for y: \(y = \frac{4}{17}\text{.}\)
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3.

Solve the system using elimination: \(\begin{cases} 2x+y-z=4\\ x-2y+3z=-6\\ 3x+y+2z=7 \end{cases}\)
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Hint.
Pick a base equation to use for eliminating the same variable in the other two equations. Solve the resulting two-variable system.
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Answer.
\(\left(\frac{3}{14}, \frac{9}{2}, \frac{13}{14}\right)\)
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Solution.
Let the second equation be the base equation. Multiply it by 2 and subtract it from the first equation to eliminate x: \(5y - 7z = 16\text{.}\) Multiply the second equation by 3 and subtract it from the third equation to eliminate x: \(7y - 7z = 25\text{.}\) Solve the resulting two-variable system for y and z. We can use elimination to find y, by subtracting the first equation from the second: \(y = \frac{9}{2}\text{.}\) Substitute this to find the value of z. \(z=\frac{13}{14}\text{.}\) Substitute these values back into the base equation to find x: \(x-2\left(\frac{9}{2}\right)+3\left(\frac{13}{14}\right)=-6\text{.}\) Solving for x we get \(x = \frac{3}{14}\text{.}\)
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4.

Find all solutions to the system: \(\begin{cases} \frac{4}{x^2}+\frac{6}{y^4}=\frac{7}{2}\\\frac{1}{x^2}-\frac{2}{y^4}=0\end{cases}\)
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Hint.
Use substitution to solve for one variable in terms of the other.
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Answer.
\(\left(-\sqrt{2}, -\sqrt{2}\right), \left(\sqrt{2}, \sqrt{2}\right), \left(-\sqrt{2}, \sqrt{2}\right), \left(\sqrt{2}, -\sqrt{2}\right)\)
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Solution.
Here we can use substitution. From the second equation, we have \(\frac{1}{x^2} = \frac{2}{y^4}\text{.}\) Substituting this into the first equation gives us \(4 \cdot \frac{2}{y^4} + \frac{6}{y^4} = \frac{7}{2}\text{.}\) Simplifying, we get \(\frac{14}{y^4} = \frac{7}{2}\text{.}\) Solving for \(y^4\text{,}\) we find \(y^4 = 4\text{,}\) so \(y = \pm \sqrt{2}\text{.}\) Substituting back into the second equation, we can find the corresponding values of \(x\text{.}\) So \(x = \pm \sqrt{2}\text{.}\)
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