3 2 Solving Systems Algebraically Substitution Elimination EXAMPLE

3. 2 Solving Systems Algebraically Substitution & Elimination

EXAMPLE 1 Use the substitution method Solve the system using the substitution method. 2 x + 5 y = – 5 Equation 1 x + 3 y = 3 Equation 2 SOLUTION STEP 1 Solve Equation 2 for x. x = – 3 y + 3 Revised Equation 2

EXAMPLE 1 Use the substitution method STEP 2 Substitute the expression for x into Equation 1 and solve for y. 2 x +5 y = – 5 2( – 3 y + 3 ) + 5 y = – 5 y = 11 Write Equation 1. Substitute – 3 y + 3 for x. Solve for y. STEP 3 Substitute the value of y into revised Equation 2 and solve for x. x = – 3 y + 3 Write revised Equation 2. x = – 3( ) + 3 Substitute 11 for y. x = – 30 Simplify.

EXAMPLE 1 Use the substitution method ANSWER The solution is (– 30, 11). CHECK Check the solution by substituting into the original equations. 2(– 30) + 5(11)= – 5? – 5 = – 5 Substitute for x and y. Solution checks. – 30 + 3(11) = 3 ? 3=3

EXAMPLE 2 Use the elimination method Solve the system using the elimination method. 3 x – 7 y = 10 Equation 1 6 x – 8 y = 8 Equation 2 SOLUTION STEP 1 Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign. – 6 x + 14 y = -20 3 x – 7 y = 10 6 x – 8 y = 8 STEP 2 Add the revised equations and solve for y. 6 y = – 12 y = – 2

EXAMPLE 2 Use the elimination method STEP 3 Substitute the value of y into one of the original equations. Solve for x. 3 x – 7 y = 10 Write Equation 1. 3 x – 7(– 2) = 10 Substitute – 2 for y. 3 x + 14 = 10 x= – Simplify. 4 3 Solve for x.

EXAMPLE 2 Use the elimination method ANSWER – 4 The solution is ( 3 , – 2 ) CHECK You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.

GUIDED PRACTICE for Examples 1 and 2 Solve the system using the substitution for 1 and the elimination method for 2. 1. 4 x + 3 y = – 2 x + 5 y = – 9 ANSWER The solution is (1, – 2). 2. 3 x + 3 y = – 15 5 x – 9 y = 3 ANSWER The solution is ( -3 , – 2)

GUIDED PRACTICE for Examples 1 and 2 Solve the system using the substitution or the elimination method. 3. 3 x – 6 y = 9 – 4 x + 7 y = – 16 ANSWER The solution is (11, 4)

EXAMPLE 4 Solve linear systems with many or no solutions Solve the linear system. a. x – 2 y = 4 b. 3 x – 6 y = 8 4 x – 10 y = 8 – 14 x + 35 y = – 28 SOLUTION a. Because the coefficient of x in the first equation is 1, use the substitution method. Solve the first equation for x. x – 2 y = 4 x = 2 y + 4 Write first equation. Solve for x.

EXAMPLE 4 Solve linear systems with many or no solutions Substitute the expression for x into the second equation. 3 x – 6 y = 8 3(2 y + 4) – 6 y = 8 12 = 8 Write second equation. Substitute 2 y + 4 for x. Simplify. ANSWER Because the statement 12 = 8 is never true, there is no solution.

EXAMPLE 4 Solve linear systems with many or no solutions b. Because no coefficient is 1 or – 1, use the elimination method. Multiply the first equation by 7 and the second equation by 2. 4 x – 10 y = 8 – 14 x + 35 y = – 28 Add the revised equations. 28 x – 70 y = 56 – 28 x + 70 y = – 56 0 = 0 ANSWER Because the equation 0 = 0 is always true, there are infinitely many solutions.

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 5. 12 x – 3 y = – 9 – 4 x + y = 3 SOLUTION Because the coefficient of y in the second equation is y, use the substitution method. Solve the 2 nd equation for y. – 4 x + y = 3 y = 4 x + 3 Write second equation. Solve for y.

GUIDED PRACTICE for Example 4 Substitute the expression for y into the first equation. 12 x – 3 y = – 9 12 x – 3(4 y + 3) = – 9 0=0 Write second equation. Substitute 4 x + 3 for y. Simplify. ANSWER Because the equation 0 = 0 is always true, there are infinitely many solutions.

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 6. 6 x + 15 y = – 12 – 2 x – 5 y = 9 Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 3 6 x + 15 y = – 12 – 2 x – 5 y = 9 Add the revised equations. 6 x + 15 y = – 12 3 – 6 x – 15 y = 27 0 = 15 ANSWER Because the statement 0 = 15 is never true, there are no solutions.

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 7. 5 x + 3 y = 20 –x– 3 y = – 4 5 Because the coefficient of x in the first equation is – 1, use the substitution method. Solve the 2 nd equation for x. –x– 3 y = – 4 5 x=– 3 y+4 5 Write second equation. Solve for x.

GUIDED PRACTICE for Example 4 Substitute the expression for x into the first equation. 5 x + 3 y = 20 5 (– 3 y + 4 ) + 3 y = 20 5 20 = 20 Write first equation. 3 y+4 Substitute – 5 for x. Simplify. ANSWER Because the statement 20 = 20 is always true, there are infinitely many solution.

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 8. 12 x – 2 y = 21 3 x + 12 y = – 4 Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 6 12 x – 2 y = 21 3 x + 12 y = – 4 Add the revised equations. 6 72 x – 12 y = 126 3 x + 12 y = – 4 75 x = 122 75

for Example 4 GUIDED PRACTICE Substitute the expression for x into the second equation. 3 x + 12 y = – 4 3( 122 ) + 12 y = 8 75 – 37 y= 50 Write second equation. 122 Substitute for x. 75 Simplify. ANSWER The solution is ( , 122 ) 75 – 37 50

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 9. 8 x + 9 y = 15 5 x – 2 y = 17 ANSWER (3, – 1) 10. 5 x + 5 y = 5 5 x + 3 y = 4. 2 ANSWER (0. 6, 0. 4)