Virtual Help
- Chat with library staff now
- Contact your library
Math: Basic Tutorials : Solving Systems of Linear Equations
Solving Systems of Linear Equations
In previous modules, you solved algebraic equations for one variable, but what if you have more than one variable? A system of linear equations consists of two or more linear equations made up of two or more variables. To solve a linear system, we need at least as many equations as there are variables. In this module, you will review how to solve systems of linear equations with two variables by graphing and using the algebraic methods of substitution and elimination.
Examples & Activity
Examples
Click on the titles below to view each example.
Solve the following linear system by graphing. 2x plus y equals negative 8 and x minus y equals negative 1.
Solution:
Line 1: Rearrange the first equation to isolate y. 2x plus y equals negative 8 is rearranged to y equals negative 2x minus 8.
Line 2: Identify the slope and the y-intercept of the first equation. The slope of the first line is 2 and the y-intercept is negative 8.
Line 3: Rearrange the second equation to isolate y. x minus y equals negative 1 is rearranged to y equals x plus 1.
Line 4: Identify the slope and the y-intercept of the second equation. The slope of the second line is 1 and the y-intercept is 1.
Line 5: Graph the lines using the slope and y-intercepts and identify the point of intersection. The first line, y equals negative 2x minus 8, crosses the y axis at negative 8 and has rise of 2 and run of 1. The second line, y equals x plus 1, crosses the y axis at 1 and has rise of 1 and a run of 1. The two lines intersect at the point (negative 3, 2).
Line 6: The solution to the linear system is (negative 3, 2).
Solve the following linear system by substitution. Negative x plus y equals negative 5 and 2x minus 5y equals 1.
Line 1: Rearrange the first equation to isolate for y. The equation negative x plus y equals negative 5 is rearranged to y equals negative 5 plus x.
Line 2: Substitute the expression negative 5 plus x for y into the second equation. After substituting, 2x minus 5y equals 1 becomes 2x minus 5 open parentheses negative 5 plus x close parentheses equals 1.
Line 3: Solve for x. To solve for x, use the distributive property, so the equation becomes 2x plus 25 minus 5x equals 1. Collect like terms and rearrange so the equation is negative 3 x equals negative 24. Divide both sides by negative 3 to get x equals 8.
Line 4: Now, substitute x equals into the first equation to solve for y. After substituting, the first equation, negative x plus y equals negative 5, becomes negative 8 plus y equals negative 5.
Line 5: Solve for y. y equals 3.
Line 6: The solution to the linear system is open parentheses 8, 3 close parentheses.
Line 7. Check your solution by substituting open parentheses 8, 3 close parentheses into the first equation. Negative 8 plus 3 equals negative 5. Negative 5 equals negative 5, so this is a true statement.
Line 8: Check your solution by substituting open parentheses 8, 3 close parentheses into the second equation. 2 times 8 minus 5 times 3 equals 1. 1 equals 1, so this is a true statement.
Line 9: The solution makes both equations true so it is correct.
Solve the following linear system by elimination. 3x plus 5y equals negative 11 and x minus 2y equals 11.
Solution:
Line 1: Multiply the second equation by negative 3, so that the numerical coefficients in front of the x are the same in both equations but have opposite signs. -3 times open parentheses x minus 2y close parenthesis equals -3 times open parenthesis 11 close parentheses becomes negative 3x plus 6y equals negative 33.
Line 2: Using the first equation and the newly created equation, add the equations to eliminate the x term. 3x plus 5y equals negative 11 added to negative 3x plus 6y equals negative 33 becomes 11y equals negative 44.
Line 3: Solve for y. y equals negative 4.
Line 4: Now, substitute y equals negative 4 into one of the original equations to solve for x. After substituting, equation 1 becomes x minus 2 times negative 4 equals 11.
Line 5: Solve for x. x plus 8 equals 11, so x equals 3.
Line 6: The solution to this system of linear equations is open parentheses 3, negative 4 close parentheses
Line 7: Check your solution by substituting open parentheses 3, negative 4 close parentheses into the first equation. 3 times 3 plus 5 times negative 4 equals negative eleven. Negative 11 equals negative 11, so this is a true statement.
Line 8: Check your solution by substituting open parentheses 3, negative 4 close parentheses into the second equation. 3 minus 2 times negative 4 equals 11. 11 equals 11, so this is a true statement.
Line 9: The solution makes both equations true so it is correct.
Summary and Worksheet
- Summary: Solving Systems of Linear Equations - PDF - Opens in a new windowThis document contains a short summary of this topic as well as detailed examples to illustrate key concepts. Use this summary to review this topic.
- Worksheet: Solving Systems of Linear Equations - PDF - Opens in a new windowThis document contains practice questions on this topic. Use the worksheet to test your knowledge and practice the skills learned in this module. The answers to the practice questions are provided at the end.
Attribution
Examples Source: "College Algebra - opens in a new window" by Jay Abramson is licensed under CC BY 4.0 - opens in a new window / A derivative from the original work - opens in a new window