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Nonamiya [84]
3 years ago
6

How do I solve this?

Mathematics
2 answers:
Leya [2.2K]3 years ago
5 0
Hi there!

Let's start by looking at our like terms:
12 and 7
2v and 7v
13v^3 and 13v^3

Since all of those terms are like each other, they can be added together. The terms that are being cubed can only be added to other terms being cubed because otherwise things would be a bit messed up. Here's what we can add together:
12 + 7 = 19
2v + 7v = 9v
13v^3 + 13v^3 = 26v^3

Now, we need to organize the terms in the expression like this - variable and cubed, just a variable, number. Using that our expression ends up being:
26v^3 + 9v + 19

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
FinnZ [79.3K]3 years ago
4 0
(12 + 2v + 13v^3) + (13v^3 + 7 + 7v)
12 + 2v + 13v^3 + 13v^3 + 7 + 7v - drop the parenthesis since adding
13v^3 + 13v^3 + 7v + 2v + 12 + 7 - Group by like terms
26v^3 + 9v + 19 - combine like terms
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The solution is (-1,5) and (7,-3)

Step-by-step explanation:

The expression is x^{2} +y^{2} -x+3y-42=0 and x+y=4

Using substitution method we can solve the expression.

Let us substitute x=4-y in x^{2} +y^{2} -x+3y-42=0

(4-y)^{2} +y^{2} -(4-y)+3y-42=0

Expanding and simplifying the expression, we get,

\begin{array}{r}{16-8 y+y^{2}+y^{2}-4+y+3 y-42=0} \\{2 y^{2}-4 y-30=0}\end{array}

Let us use the quadratic equation formula to solve this equation,

\begin{aligned}y &=\frac{4 \pm \sqrt{16-4(2)(-30)}}{2(2)} \\&=\frac{4 \pm \sqrt{16+240}}{4} \\&=\frac{4 \pm 16}{4} \\y &=1 \pm 4\end{aligned}

Thus, y=5 and y=-3

Substituting y-values in the equation x+y=4, we get the value of x.

For y=5 ⇒ x=4-5=-1

For y=-3 ⇒ x=4+3=7

Thus, the solution set is (-1,5) and (7,-3)

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3 years ago
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Step-by-step explanation:

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