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Chapter 8.4. Review of Methods for Solving Systems of Equations. 1a. Select a method to solve. 3x + 5 y = 1485. x + 2y = 564. Let’s review substitution. 1a. Select a method to solve. 3. x. + 5 y. x. = 1485. + 2y = 564. -2y -2y. x =. -2y + 564. step 1 solve for x.
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Chapter 8.4 Review of Methods for Solving Systems of Equations
1a. Select a method to solve. 3x + 5y = 1485 x + 2y = 564 Let’s review substitution.
1a. Select a method to solve. 3 x +5y x = 1485 + 2y = 564 -2y -2y x = -2y + 564 step 1 solve for x
1a. Select a method to solve. 3 x +5y x = 1485 + 2y = 564 -2y -2y (-2y + 564) 3 + 5y = 1485 x = -2y + 564 step 2 substitute into x
1a. Select a method to solve. 3 x +5y x = 1485 + 2y = 564 -2y -2y (-2y + 564) 3 + 5y = 1485 x = -2y + 564 -6y +5y = 1485 + 1692 -y = 1485 + 1692 -1692 -1692 -y = -207 -1 -1 y = 207 step 3 solve for y
1a. Select a method to solve. 3 x +5y x = 1485 + 2y = 564 -2y -2y (-2y + 564) 3 + 5y = 1485 x = -2y + 564 -6y +5y = 1485 + 1692 x = -2(207) + 564 -y = 1485 + 1692 -1692 -1692 x = -414 + 564 -y = -207 x = 150 -1 -1 y = 207 (150, 207) step 4 substitute and solve for x
1b. Select a method to solve. -5y = -2(3x + 1) 7x – 3 = -6(y – 7) Let’s review elimination.
1b. Select a method to solve. -5y = -2(3x + 1) 7x – 3 = -6(y – 7) -6y + 42 7x – 3 = -6x – 2 -5y = +6y +6y +6x +6x +3 +3 7x 6x + 6y – 5y = 45 = -2 Before we begin let’s rewrite each equation in the form Ax + By = C. Use these equations to solve by elimination.
1b. Select a method to solve. = 45 + 6y 7x = -2 – 5y 6x step 1 coefficients of one variable must be opposite.
1b. Select a method to solve. = 45 + 6y 7x 5( ) = -2 – 5y 6x 6( ) 35x + 30y = 225 36x – 30y = -12 step 2 make the y opposites, multiply first equation by 5 and second equation by 6.
1b. Select a method to solve. = 45 + 6y 7x 5( ) = -2 – 5y 6x 6( ) 35x + 30y = 225 36x – 30y = -12 71x = 213 step 3 add to eliminate the y.
1b. Select a method to solve. = 45 + 6y 7x 5( ) = -2 – 5y 6x 6( ) 35x + 30y = 225 36x – 30y = -12 71x = 213 71 71 x = 3 step 4 solve for x.
1b. Select a method to solve. = 45 + 6y 7x 5( ) 7(3) + 6y = 45 = -2 – 5y 6x 6( ) 21 + 6y = 45 -21 -21 35x + 30y = 225 6y = 24 36x – 30y = -12 6 6 71x = 213 71 71 y = 4 x = 3 (3, 4) step 5 substitute into equation 1 and solve for y.
2. Solve algebraically. 4x + 2y = 2 -6x – 3y = 6 Solve by elimination. step 1 coefficients of one variable must be opposite.
2. Solve algebraically. 4x + 2y 3( ) = 2 -6x – 3y = 6 2( ) 12x + 6y = 6 -12x – 6y = 12 step 2 make the x opposites, multiply first equation by 3 and second equation by 2.
2. Solve algebraically. 4x + 2y 3( ) = 2 -6x – 3y = 6 2( ) 12x + 6y = 6 -12x – 6y = 12 0 18 ≠ No solution. The system is inconsistent. Both x and y are eliminated and 0 is not equal to 18. step 3 add to eliminate the x.
3. Solve algebraically. 3x – 9y = 18 -4x + 12y = -24 Solve by elimination. step 1 coefficients of one variable must be opposite.
3. Solve algebraically. 4( ) 3x – 9y = 18 3( ) -4x + 12y = -24 12x – 36y = 72 -12x + 36y = -72 step 2 make the x opposites, multiply first equation by 4 and second equation by 3.
3. Solve algebraically. 4( ) 3x – 9y = 18 3( ) -4x + 12y = -24 12x – 36y = 72 -12x + 36y = -72 0 = 0 Infinite number of solutions. The equations are dependent. Both x and y are eliminated and 0 is equal to 0. step 3 add to eliminate the x.
Chapter 8.4 Review of Methods for Solving Systems of Equations