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anonymous
 one year ago
HELP! WILL MEDAL AND FAN!!!!
A system of equations is shown below:
3x + 7y = 16
9x + 5y = 16
Part A: Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this. (6 points)
Part B: Show that the equivalent system has the same solution as the original system of equations.
anonymous
 one year ago
HELP! WILL MEDAL AND FAN!!!! A system of equations is shown below: 3x + 7y = 16 9x + 5y = 16 Part A: Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this. (6 points) Part B: Show that the equivalent system has the same solution as the original system of equations.

This Question is Closed

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Part A: You can start this way: Multiply the first equation by 2 on both sides. Now add it to the second equation. What do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03x + 7y = 16 X2 X2

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Remember to distribute the 2 on the left side. \(2 \times (3x + 7y) = 2 \times (16) \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i solve this or something?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1No, not yet. We are working on multiplying the first equation by 2 on both sides. On the left side, you need to distribute the 2. See below. \(\color{red}{2 \times} (\color{green}{3x} + \color{blue}{7y}) = 2 \times (16)\) \(\color{red}{2 \times } \color{green}{3x} + \color{red}{2 \times } \color{blue}{7y} = 32\) \(6x + 14y = 32\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1So far, we have multiplied the first equation by 2 on both sides. Now we add that new equation to the second original equation: \(~~~~~~~~~~6x + 14y = 32\) \(~~~+~~~9x + ~~5y = 16\)  \(~~~~~~~~~~15x + 19y = 16\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are we still on Part A?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont have a lot of time and i still have 2 more questions to do after this one

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1The new equation just above is the sum of one original equation and a multiple of the other original equation. Now we take the new equation and one of the original equations, and that completes Part A. 3x + 7y = 16 15x + 19y = 16

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1That is the end of part A. We have a new system that is equivalent to the original one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so what should i type in for part a?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Everything we did above.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Skipping the explanations.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can it be put into not that because i cant write like that in my assignment

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Part A. 3x + 7y = 16 9x + 5y = 16 2(3x + 7y) = 2(16) 6x + 14y = 32 + 9x + 5y = 16  15x + 19y = 16 3x + 7y = 16 15x + 19y = 16

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Ok. The response above is Part A.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay thank you can you help me with part b too?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Yes. Now in Part B, we need to show both systems are equivalent to each other and both have the same solution. We could solve both systems, but there is a faster way. We solve only the original system. Then we try the solution of the original system in the second system. If the solution works, that shows the systems are equivalent.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Let's solve the original system. We can use the addition method. We need to add the two equations together and make sure one of the variables adds to zero.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Since the first term of the first equation is 3x, if we multiply it by 3, we get 9x. 9x added to the first term of the second equation, 9x, will add to zero. That is what we want, so we start by multiplying the first equation by 3 on both sides. 3x + 7y = 16  *3 on both sides  > 9x 21y = 48 9x + 5y = 16  copied just as it is > 9x + 5y = 16   sum of the two equations > 16y = 64 16y = 64 y = 4 Now that we know y = 4, we substitute 4 in the first equation for y. Then we find x. 3x + 7y = 16 3x + 7(4) = 16 3x 28 = 16 3x = 12 x = 4 x is also 4. The solution of the original system of equations is: x = 4, y = 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that is our part b correct?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1One more step. We need to check the solution we got for the first system in the second system.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1We can solve this system, but it's faster to just check the solution we already know. The question now is: Is the solution x = 4 and y = 4 also a solution to the new system of equations? Let's test (4, 4) in both equations of the second system. 3x + 7y = 16 3(4) + 7(4) =? 16 12 + (28) =? 16 16 = 16 The solution works on the first equation. 15x + 19y = 16 15(4) + 19(4) =? 16 60 + (76) =? 16 60  76 =? 16 16 = 16 The solution works on the second equation. This just showed that the two systems are equivalent. Part B is done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i get it thanks so much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u help with another? @mathstudent55

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1You're welcome. I'll look.
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