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So far, you have mostly been solving for a single variable in an equation or placing one variable in terms of the others. However, you will likely face questions asking you to solve for more than one variable, and these questions will likely be about linear equations. A linear equation simply refers to an equation that will result in a single straight line on a graph, such as slope‐intercept y = mx + b. To solve for more than one variable, you will be working a system of linear equations. A system is simply a collection of equations that are solved at the same time and come to the same result. This means the only solution to all equations within the system will be the intersection of all the lines on the graph, or where they share the same point. Make sure to have a solid foundation of solving one-variable equations before attempting systems of linear equations. You can solve a system through elimination or substitution, though both methods are variations of one another. Solving One-Variable Equations This is a simple one-variable algebraic equation. The objective is to isolate the variable, x on one side of the equation. 3x - 7= 11 Remember that equations must be balanced on both sides so what you do on one side you do to the other. Add 7 to both sides of the equation to get 3x by itself. 3x = 18 Solve for x by dividing each side by 3. x = 6 Solving Systems through Elimination Eliminate one variable from the system by manipulating one or both equations so that the sum of one variable equals zero. 2x + y = 5 -5x - 4y = 4 Add the two equations together. Do any of the coefficients of the variable equal zero? If not, choose to eliminate either variable, but let’s start by making the y’s in both equations sum to zero. If we make the y in the first equation 4y, then the sum of the y’s in the first and second equations will equal zero. To make this happen, multiply the entire first equation by 4. 4 (2x + y) = 4(5) Let’s add the two equations. Since y equals zero, we can omit y from the new equation. Solve for x. 3x = 24 → x = 8 Pick an equation- either should work for this step. Plug in the value for x into either equation and solve for y. 2(8) + y = 5 16 + y = 5 y = -11 Check your work by putting the values for x and y into the remaining equation. If both equations are equal, then your answers for x and y are correct. Solving Sytems through Substitution When it’s not readily apparent how you should manipulate the equations, you can use substitution. Select the simplest equation. Since the coefficient for y = 1, let’s set the equation equal to y. Now that the equation is written in terms of y, substitute this equation into the second equation instead of the value of y. At this point, x should be the only variable to solve for. Solve for x by completing the arithmetic above to find that: Plug in the value for x into either of the two equations and solve for y. 2(8) + y = 5 → 16 + y = 5 y = -11 Check your work by plugging the values for x and y into both equations. If both equations are equal, then the answer is correct! -5(8) - 4(-11) = -40 + 44 = 4 You will notice that our substitution efforts resulted in the same equation with one variable that our elimination efforts did. The only difference is that the 20 was already added to the other side in our elimination efforts. To solve these elimination and substitution problems quickly, choose the simplest equations with the lowest coefficient values as these will take less math to solve. You should have a good understanding of the distributive property before attempting this problem. Practice Problems Answers to Practice Problems 2 1 9/7 1 3.5
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