12 Systems of Linear Equations and Inequalities

12.2 Solving Systems of Linear Equations by Substitution Objectives 1. Solve linear systems by substitution. 2. Solve special systems by substitution. 3. Solve linear systems with fractions and decimals. 2

Solve Linear Systems by Substitution Example 1 Solve the system by the substitution method. 5x + 2y = 2 y = – 3x The second equation is already solved for y. This equation says that y = – 3x. Substituting – 3x for y in the first equation gives 5x + 2y = 2 5x + 2(– 3x) = 2 Let y = – 3x. 5x + – 6x = 2 Multiply. – 1x = 2 Combine like terms. x = – 2 Multiply by – 1.

Solve Linear Systems by Substitution Example 1 (cont.) Because x = – 2, we find y from the equation y = – 3x by substituting – 2 for x. y = – 3x = – 3(– 2) = 6 Let x = – 2. Check that the solution of the given system is (– 2, 6) by substituting – 2 for x and 6 for y in both equations. 5x + 2y = 2 y = – 3x 5(– 2) + 2(6) = 2 ? 6 = – 3(– 2) ? 2 = 2 True 6 = 6 True The solution set of the system is {(– 2, 6)}.

Solve Linear Systems by Substitution Example 2 Solve the system by the substitution method. 3x + 4y = 4 x = 2y + 18 The second equation gives x in terms of y. Substitute 2y + 18 for x in the first equation. 3x + 4y = 4 3(2y + 18) + 4y = 4 Let x = 2y + 18. 6y + 54 + 4y = 4 Distributive property 10y + 54 = 4 Combine like terms. 10y = – 50 Subtract 54. y = – 5 Divide by 10.

Solve Linear Systems by Substitution Example 2 (continued) Because y = – 5, we find x from the equation x = 2y + 18 by substituting – 5 for y. x = 2y + 18 = 2(– 5) + 18 = 8 Let y = – 5. Check that the solution of the given system is (8, – 5) by substituting 8 for x and – 5 for y in both equations. 3x + 4y = 4 x = 2y + 18 3(8) + 4(– 5) = 4 ? 8 = 2(– 5) + 18 ? 4 = 4 True 8 = 8 True The solution set of the system is {(8, – 5)}.

Solve Linear Systems by Substitution Solving a Linear System by Substitution Step 1 Solve one equation for either variable. If one of the variables has a coefficient of 1 or – 1, choose it, since it usually makes the substitution easier. Step 2 Substitute for that variable in the other equation. The result should be an equation with just one variable. Step 3 Solve the equation from Step 2. Step 4 Substitute the result from Step 3 into the equation from Step 1 to find the value of the other variable. Step 5 Check the solution in both of the original equations. Then write the solution set.

Solve Linear Systems by Substitution Example 3 Solve the system by the substitution method. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 1 For the substitution method, we must solve one of the equations for either x or y. Because the coefficient of y in equation (2) is – 1, we choose equation (2) and solve for y. 5x – y = 13 (2) 5x – y – 5x = 13 – 5x Subtract 5x. – y = 13 – 5x Combine like terms. y = – 13 + 5x Multiply by – 1.

Solve Linear Systems by Substitution Example 3 (continued) Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 2 Now substitute – 13 + 5x for y in equation (1). 3x + 2y = 26 (1) 3x + 2(– 13 + 5x) = 26 Let y = – 13 + 5x.

Solve Linear Systems by Substitution Example 3 (continued) Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 3 Now solve the equation from Step 2. 3x + 2(– 13 + 5x) = 26 From Step 2. 3x – 26 + 10x = 26 Distributive property 13x – 26 = 26 Combine like terms. 13x – 26 + 26 = 26 + 26 Add 26. 13x = 52 Combine like terms. x = 4 Divide by 13.

Solve Linear Systems by Substitution Example 3 (continued) Use substitution to solve the system. 3x + 2y = 26 (1) 5x – y = 13 (2) Step 4 Since y = – 13 + 5x and x = 4, y = – 13 + 5(4) = 7. Step 5 Check that (4, 7) is the solution. 3x + 2y = 26 5x – y = 13 3(4) + 2(7) = 26 ? 5(4) – 7 = 13 ? 12 + 14 = 26 ? 20 – 7 = 13 ? 26 = 26 True 13 = 13 True Since both results are true, the solution set of the system is {(4, 7)}.

Solve Special Systems by Substitution Example 4 Use substitution to solve the system. y = 2x – 7 (1) 3y – 6x = –6 (2) Substitute 2x – 7 for y in equation (2). 3y – 6x = –6 (2) 3(2x – 7) – 6x = –6 Let y = 2x – 7. 6x – 21 – 6x = –6 Distributive property False. – 21 = –6 This false result means that the equations in the system have graphs that are parallel lines. The system is inconsistent and the solution set is

Solve Special Systems by Substitution x y (2) (1) Example 5 (continued) y = 2x – 7 (1) 3y – 6x = – 6. (2)

Solve Special Systems by Substitution Example 5 Solve the system by the substitution method. 3 = 5x – y (1) – 4y – 12 = – 20x (2) Begin by solving equation (1) for y to get y = 5x – 3. Substitute 5x – 3 for y in equation (2) and solve the resulting equation. – 4y – 12 = – 20x (2) – 4(5x – 3) – 12 = – 20x Let y = 5x – 3. – 20x + 12 – 12 = – 20x Distributive property 0 = 0 Add 20x; combine terms. This true result means that every solution of one equation is also a solution of the other, so the system has an infinite number of solutions.

Solve Special Systems by Substitution x y Example 5 (continued) 3 = 5x – y (1) – 4y – 12 = – 20x (2) The system has an infinite number of solutions – all the ordered pairs corresponding to points that lie on the common graph.

Solve Linear Systems with Fractions and Decimals Example 6 Solve the system by the substitution method. x + 5y = 1 2 3 (1) x – y = – 1 4 8 13 (2) Clear equation (1) of fractions by multiplying each side by 3. x + 5y = 1 2 3 Multiply by 3. x + 5y = 1 2 3 Distributive property 2x + 15y = 3 (3)

Solve Linear Systems with Fractions and Decimals Example 6 (cont.) Solve the system by the substitution method. x + 5y = 1 2 3 (1) x – y = – 1 4 8 13 (2) Now, clear equation (2) of fractions by multiplying each side by 8. x – y = – 1 4 8 13 Multiply by 8. x – y = – 1 4 8 13 Distributive property 2x – y = – 13 (4)

Solve Linear Systems with Fractions and Decimals Example 6 (cont.) The given system of equations has been simplified to the equivalent system. (1) (2) x + 5y = 1 2 3 x – y = – 1 4 8 13 2x – y = – 13 2x + 15y = 3 (3) (4) To solve this system by substitution, equation (4) can be solved for y. 2x – y = – 13 (4) 2x – y – 2x = – 13 – 2x Subtract 2x. – y = – 13 – 2x Combine like terms. y = 13 + 2x Multiply by – 1.

Solve Linear Systems with Fractions and Decimals Example 6 (cont.) The given system of equations has been simplified to the equivalent system. (1) (2) x + 5y = 1 2 3 x – y = – 1 4 8 13 2x – y = – 13 2x + 15y = 3 (3) (4) Now substitute 13 + 2x for y in equation (3). 2x + 15y = 3 (3) 2x + 15(13 + 2x) = 3 Let y = 13 + 2x. 2x + 195 + 30x = 3 Distributive property 32x + 195 = 3 Combine like terms.

Solve Linear Systems with Fractions and Decimals Example 6 (cont.) The given system of equations has been simplified to the equivalent system. (1) (2) x + 5y = 1 2 3 x – y = – 1 4 8 13 2x – y = – 13 2x + 15y = 3 (3) (4) Now substitute 13 + 2x for y in equation (3). 32x + 195 = 3 Combine like terms. 32x + 195 – 195 = 3 – 195 Subtract 195. 32x = – 192 Combine like terms. x = – 6 Divide by 32.

Solve Linear Systems with Fractions and Decimals Example 6 (cont.) The given system of equations has been simplified to the equivalent system. (1) (2) x + 5y = 1 2 3 x – y = – 1 4 8 13 2x – y = – 13 2x + 15y = 3 (3) (4) Substitute – 6 for x in y = 13 + 2x to get y = 13 + 2(– 6) = 1. Check by substituting – 6 for x and 1 for y in both of the original equations. The solution set is {(– 6, 1)}.