Algebra I : Semester II : Solving Systems

Sections:

Introduction  |   Section 1  |   Section 2  |   Section 3  |  Section 4  |  Section 5

Section Two

Part 1  |  Part 2  |  Part 3  |  Part 4

Algebra 1: Section 2: Solving Systems Solving With Substitution   Click here for a presentation on the Substitution Method. Now let’s put some of the things we found out about the substitution method in a form to remember.   The Substitution Method Write one of the equations in terms of x or y. Substitute the expression on the right of the equals sign into the variable in the other equation. Solve for the variable. Substitute the value into the original equation and solve for the other variable. Write the solution as an ordered pair, (x, y). Example 1: Solve Using Substitution y = x + 5 28x + y = 5 Write one of the equations in terms of x or y.   y = x + 5 The first equation is already in terms of y. Substitute the expression on the right into the variable in the other equation.   28x + y = 5 28x + (x + 5) = 5   Solve for the variable.   28x + x + 5 = 5 29x + 5 = 5 29x = 5 – 5 29x = 0 x = 0/29 x = 0 Simplify Simplify Subtraction Property Simplify Division Property Simplify Substitute the value into your original equation and solve for the other variable.   y = x + 5 y = 0 + 5 y = 5 Substitute x = 0 Write solution as an ordered pair (x,y) (0, 5)   Check the solution: y = x + 5 5 = 0 + 5 5 = 5  √ 28x + y = 5 28(0) + 5 = 5 0 + 5 = 5 5 = 5   √ Therefore, (0, 5) is the solution of the system. Example 2:  Solve for a Variable First Solve the system. 5x – 2y = 4 4x – y = 5 Write one of the equations in terms of x or y.   4x – y = 5 -y = 5 – 4x y = -5 + 4x Choose the equation where x or y has a coefficient of 1 or -1, if possible. Substitute the expression on the right into the variable in the other equation.   5x – 2y = 4 5x – 2(-5 + 4x) = 4     Solve for the variable.   5x + 10 – 8x = 4 -3x + 10 = 4 -3x = 4 – 10  -3x = -6 x = -6/-3 x = 2 Distribute -2 Simplify Subtraction Property Simplify Division Property Simplify Substitute the value into your original equation and solve for the other variable.   y = -5 + 4x y = -5 + 4(2) y = -5 + 8 y = 3 Substitute x = 2 Write the solution as an ordered pair, (x, y). (2, 3)   Check the solution: 5x – 2y = 4 5(2) – 2(3) = 4 10 – 6 = 4 4 = 4  √ 4x – y = 5 4(2) – 3 = 5 8 – 3 = 5 5 = 5   √ Therefore, (2, 3) is the solution of the system. Example 3:  A System with No Solution Solve the system. 2x + 6y = 2 x + 3y = 8 Write one of the equations in terms of x or y.   x + 3y = 8 x = 8 – 3y Choose the equation where x or y has a coefficient of 1 or -1, if possible. Substitute the expression on the right into the variable in the other equation.   2x + 6y = 2 2(8 – 3y) + 6y = 2   Solve for the variable.   16 – 6y + 6y = 2 16 = 2 This is a false statement! Distribute 2 Simplify If we have a false statement while solving, the system has no solution. No Solution   Example 4:  A System with Infinitely Many Solutions Solve the system. x – y = 10 3x – 3y = 30 Write one of the equations in terms of x or y.   x – y = 10 x = 10 + y Choose the equation where x or y has a coefficient of 1 or -1, if possible. Substitute the expression on the right into the variable in the other equation.   3x – 3y = 30 3(10 + y) – 3y = 30   Solve for the variable.   30 + 3y – 3y = 30 30 = 30 This is always true regardless of the value of x or y! Distribute 3 Simplify If we have a true statement while solving, the system has infinite solutions.  The two equations are the same equation. Infinitely many solutions   Quick Practice Solve each of the following using substitution. System Solution Details y = 3x 4x + 2y = 30     x + y = 6 x – y = 2     x + 3y = 4 3x - y = -18     x – 3y = 3 3x – 9y = 11     4x + 2y = 10 2x + y = 5     System Solution Details y = 3x 4x + 2y = 30 (3, 9)  Use y = 3x 4x + 2(3x) = 30       4x + 6x = 30 10x = 30 x = 3 y = 3x y = 3(3) y = 9   x + y = 6 x – y = 2 (4, 2) Solve the first equation for y. y = 6 – x x – (6 – x) =2      x – 6 + x = 2 2x – 6 = 2 2x = 8  x = 4 y = 6 –x y = 6 – 4 y = 2 x + 3y = 4 3x –y = -18 (-5, 3) Solve the first equation for x. x = 4 – 3y 3(4 – 3y) – y = -18 12 – 9y – y = -18 12 – 10y = -18 -10y = -30 y = 3 x = 4 – 3y x = 4 – 3(3) x = 4 – 9 x = -5 x – 3y = 3 3x – 9y = 11 no solution Solve the first equation for x. x = 3 + 3y 3(3 + 3y) – 9y = 11 9 + 9y – 9y = 11 9 = 11 False! 4x + 2y = 10 2x + y = 5 infinite solutions Solve the second equation for y. y = 5 – 2x 4x + 2(5 – 2x) = 10 4x + 10 – 4x = 10 10 = 10 Always true! Now go on to the next part

© 2006 Aventa Learning. All rights reserved.