3 Equation 1 Equation 2 Revised Equation 2 EXAMPLE 1 Use the substitution methodSolve the system using the substitution method.2x + 5y = –5Equation 1x + 3y = 3Equation 2SOLUTIONSTEP 1Solve Equation 2 for x.x = –3y + 3Revised Equation 2
4 Write revised Equation 2. EXAMPLE 1Use the substitution methodSTEP 2Substitute the expression for x into Equation 1 and solve for y.2x +5y = –5Write Equation 1.2( ) + 5y = –5–3y + 3Substitute –3y + 3 for x.y = 1111Solve for y.STEP 3Substitute the value of y into revised Equation 2 and solve for x.x = –3y + 3Write revised Equation 2.x = –3( ) + 3Substitute 11 for y.x = –30Simplify.
5 Substitute for x and y. Solution checks. EXAMPLE 1 Use the substitution methodThe solution is (– 30, 11).ANSWERCHECK Check the solution by substituting into the original equations.2(–30) + 5(11) –5=?Substitute for x and y.=?–30 + 3(11) 3–5 = –5Solution checks.3 = 3
6 Equation 1 Equation 2 EXAMPLE 2 Use the elimination method Solve the system using the elimination method.3x – 7y = 10Equation 16x – 8y = 8Equation 2SOLUTIONSTEP 1Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign.–6x + 14y = -203x – 7y = 106x – 8y = 86x – 8y = 8STEP 26y = –12Add the revised equations and solve for y.y = –2
7 Write Equation 1. Substitute –2 for y. Simplify. Solve for x. EXAMPLE 2Use the elimination methodSTEP 3Substitute the value of y into one of the original equations. Solve for x.3x – 7y = 10Write Equation 1.3x – 7(–2) = 10Substitute –2 for y.3x + 14 = 10Simplify.x =43–Solve for x.
8 EXAMPLE 2Use the elimination methodThe solution is ( , –2)43–ANSWERCHECKYou can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.
9 GUIDED PRACTICEfor Examples 1 and 2Solve the system using the substitution for 1 and the elimination method for 2.1.4x + 3y = –2x + 5y = –92.3x + 3y = –155x – 9y = 3The solution is (1,–2).ANSWERThe solution is ( -3 , –2)ANSWER
10 GUIDED PRACTICEfor Examples 1 and 2Solve the system using the substitution or the elimination method.3.3x – 6y = 9–4x + 7y = –16ANSWERThe solution is (11, 4)
11 Write first equation. Solve for x. EXAMPLE 4 Solve linear systems with many or no solutionsSolve the linear system.a.x – 2y = 43x – 6y = 8b.4x – 10y = 8– 14x + 35y = – 28SOLUTIONa.Because the coefficient of x in the first equation is 1, use the substitution method.Solve the first equation for x.x – 2y = 4Write first equation.x = 2y + 4Solve for x.
12 Write second equation. Substitute 2y + 4 for x. Simplify. EXAMPLE 4 Solve linear systems with many or no solutionsSubstitute the expression for x into the second equation.3x – 6y = 8Write second equation.3(2y + 4) – 6y = 8Substitute 2y + 4 for x.12 = 8Simplify.Because the statement 12 = 8 is never true, there is no solution.ANSWER
13 EXAMPLE 4Solve linear systems with many or no solutionsb.Because no coefficient is 1 or – 1, use the elimination method.Multiply the first equation by 7 and the second equation by 2.4x – 10y = 828x – 70y = 56– 14x + 35y = – 28– 28x + 70y = – 56Add the revised equations.0 = 0ANSWERBecause the equation 0 = 0 is always true, there are infinitely many solutions.
14 Write second equation. Solve for y. GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method.5.12x – 3y = – 9– 4x + y = 3SOLUTIONBecause the coefficient of y in the second equation is y, use the substitution method.Solve the 2nd equation for y.– 4x + y = 3Write second equation.y = 4x + 3Solve for y.
15 Write second equation. Substitute 4x + 3 for y. Simplify. GUIDED PRACTICEfor Example 4Substitute the expression for y into the first equation.12x – 3y = – 9Write second equation.12x – 3(4y + 3) = – 9Substitute 4x + 3 for y.0 = 0Simplify.Because the equation 0 = 0 is always true, there are infinitely many solutions.ANSWER
16 GUIDED PRACTICEfor Example 4Solve the linear system using any algebraic method.6.6x + 15y = – 12– 2x – 5y = 9Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 36x + 15y = – 126x + 15y = – 12– 2x – 5y = 93– 6x – 15y = 270 = 15Add the revised equations.ANSWERBecause the statement 0 = 15 is never true, there are no solutions.
17 Write second equation. Solve for x. GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method.7.5x + 3y = 20– x – y = – 435Because the coefficient of x in the first equation is – 1, use the substitution method.Solve the 2nd equation for x.– x – y = – 435Write second equation.x = – y + 435Solve for x.
18 Write first equation. Substitute for x. Simplify. GUIDED PRACTICE for Example 4Substitute the expression for x into the first equation.5x + 3y = 20Write first equation.5( – y + 4 ) + 3y = 203Substitute for x.53– y + 420 = 20Simplify.Because the statement 20 = 20 is always true, there are infinitely many solution.ANSWER
19 GUIDED PRACTICEfor Example 4Solve the linear system using any algebraic method.8.12x – 2y = 213x + 12y = – 4Because no coefficient is 1 or – 1, use the elimination method.Multiply the second equation by 612x – 2y = 21672x – 12y = 1263x + 12y = – 43x + 12y = – 4Add the revised equations.75x = 122x12275=
20 Write second equation. 122 Substitute for x. 75 Simplify. GUIDED PRACTICEfor Example 4Substitute the expression for x into the second equation.3x + 12y = – 4Write second equation.3( ) + 12y = 812275Substitute for x.12275y =– 3750Simplify.The solution is ( , )ANSWER12275– 3750
21 GUIDED PRACTICEfor Example 4Solve the linear system using any algebraic method.5x + 5y = 510.5x + 3y = 4.29.8x + 9y = 155x – 2y = 17(3, –1)ANSWER(0.6, 0.4)ANSWER