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dybincka [34]
3 years ago
7

Need help for 16) and 19) Solve each equation and check. Show all work please

Mathematics
2 answers:
Levart [38]3 years ago
8 0

Answer:

16) k = 50

19) r = 33

explanation:

16) -5 squared is 25

and then the two multiplies that 25 which equals 50

19) 121/11 = 11, which is then multiplied by 3 is 33

Paladinen [302]3 years ago
4 0

Step-by-step explanation:

You said the other one would be the last haha just kidding I'm glad to help.

16. \frac{k}{2}=(-5)^2

First, get rid of that parenthesis.

\frac{k}{2}=25

Now multiply both sides by 2 so that you can isolate k

2(\frac{k}{2})=25*2

k=50

19. \frac{r}{3}=\frac{121}{11}

This is a pretty easy one. If you didn't know, 121/11 is actually 11 :)

\frac{r}{3}=11

Simply multiply by 3 to isolate r :)

3(\frac{r}{3})=11*3

r=33

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Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

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Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

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To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

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Plug into x-y=2---> x-3=2---> x=5

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Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

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First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

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Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

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