Southern New Hampshire University GZ Final - 8-2 Problem Set_ Module Eight.pdf

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10/18/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 1/19 [PRINT] MAT-140-J1898 20EW1 Precalculus , 8-2 Problem Set: Module Eight Geoffrey Zimmermann, 10/18/20 at 5:07:07 PM EDT Question1: Score 4/4 Solve the system of equations by any method.−2x+ 5y= − 217x+ 2y= 15Enter the exact answer as an ordered pair, (x,y)If there is no solution, enter NS. If there is an infinite number of solutions, enter the generalsolution as an ordered pair in terms of xInclude a multiplication sign between symbols. For example,a*x.Your response . .
10/18/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 2/19 7(−2 x + 5 y ) = 7(−21) Multiply both sides by 7 . −14 x + 35 y = − 147 Use the distributive property. 2(7 x + 2 y ) = 2(15) Multiply both sides by 2 . 14 x + 4 y = 30 Use the distributive property. Now, let’s add them. −14 x + 35 y = − 147 14 x + 4 y = 30 39 y = − 117 y = − 3 For the last step, we substitute y = − 3 into one of the original equations and solve for x . −2 x + 5 y = − 21 −2 x + 5(−3) = − 21 −2 x − 15 = − 21 −2 x = − 6 x = 3 Our solution is the ordered pair (3, − 3) . Check the solution in the original second equation.
10/18/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 3/19 7 x + 2 y = 15 7(3) + 2(−3) = 15 21 − 6 = 15 True Question2: Score 4/4 Solve the system of equations by any method. 6 x + 11 y = 15 x + 2 y = 4 Enter the exact answer as an ordered pair, ( x , y ) . If there is no solution, enter NS. If there is an infinite number of solutions, enter the general solution as an ordered pair in terms of x . Include a multiplication sign between symbols. For example, a * x . Your response Correct response (-14,9) (-14,9) Auto graded Grade: 1/1.0 A+ 100% Total grade: 1.0×1/1 = 100% Feedback: In this case we use substitution. First, we will solve the second equation for x .
10/18/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 4/19 x + 2 y = 4 x = − 2 y + 4 Now we can substitute the expression −2 y + 4 for x in the first equation. 6 x + 11 y = 15 6(−2 y + 4) + 11 y = 15 −12 y + 24 + 11 y = 15 y = − 9 y = 9 Now, we substitute y = 9 into the second equation and solve for x . x + 2(9) = 4 x + 18 = 4 x = − 14 Our solution is (−14, 9) . Check the solution by substituting (−14, 9) into both equations.
10/18/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 5/19 6 x + 11 y = 15 6(−14) + 11(9) = 15 −84 + 99 = 15 True x + 2 y = 4 (−14) + 2(9) = 4 −14 + 18 = 4 True Question3: Score 4/4 Solve the system of equations by any method.

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