Lesson 3: Solving Linear Systems by Substitution: Explore 2

Module 7: Systems of Linear Systems

Explore 2

Read

Go to the textbook in order to see how a system of equations is solved by using the method of substitution.As you read through the solution, see if you can figure out how to choose which equation to rearrange and which variable to isolate. Also, once you obtain one of the values (x or y), how do you choose an equation to solve for the other variable?

Foundations and Pre-calculus Mathematics 10 (Pearson)

Read “Example 1: Solving a Linear System by Substitution” on pages 418 and 419. In what situations would Method 2, shown in the example, come in handy?

You have now seen how the method of substitution can be applied. While there are different approaches, the idea remains the same: reduce the system to a single equation that can be solved for an unknown variable.

While the systems may look different each time, you will always be able to solve them using the substitution method if a variable can be isolated in one equation and substituted into the other equation. In the next example, you will encounter a linear system that looks different from what you have seen so far in this module.
Can we come up with a way to use the substitution method to determine the solution?

Work through the next example to see one way that you can solve the system by first establishing an equivalent system with “more friendly” coefficients. You will explore equivalent systems more in the next lesson.

Example 2 Solve the linear system and then verify your solution.

Solution 2

Step 1: Clear the denominators by multiplying each equation by the constant corresponding to the lowest common multiple of the denominators. In this case, multiply the first equation by 4 and the second equation by 5.

The equivalent system is
6x - y = 40
5x + 3y = 110

Tada..... now we can solve this question like the others we have done...

Step 2: Isolate y in the equation (1), and substitute the resulting expression into equation (2).

Sub 6x - 40 into y in equation (2)

Step 3: Substitute x = 10 into the expression for y to find y.

Step 4: Verify the solution by substituting x = 10 and y = 20 into the original equations of the linear system.

Since both equations are true, ( 10, 20) or x = 10 and y = 20 is the solution.

Watch and Listen

Use the “Solving Systems by Substitution” vidoe to check your understanding of the principles of substitution as applied to solving linear systems.

Try This 10

Complete the following in your course folder ( binder).
Foundations and Pre-calculus Mathematics 10 (Pearson)
TT 10. Complete “Exercises” questions 4.a), 4.c), 5.a), 5.c), 7, 8.a), 11, 19.a), and 19.c) on pages 425 and 426.
Use the link below to check your answers to Try This 10.
Possible TT10 Solutions