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ololo11 [35]
4 years ago
8

Solve x+4y=15 and 5x-2y=9

Mathematics
1 answer:
avanturin [10]4 years ago
6 0

Answer:

Step-by-step explanation:

10x+4y=18

- X+4y=15

9x=3

X=1/3

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Answer:

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Step-by-step explanation:

So I believe the problem is this:

\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}

where we are asked to find values for a and b such that the equation holds for any x in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)

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As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

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Divide both sides by -10:

\frac{-50}{-10}=b

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Now we have:

2(4x+3)=a(x+7)+b(x-3) with b=5

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

2(4\cdot 3+3)=a(3+7)+b(3-3)

2(12+3)=a(10)+b(0)

2(15)=10a+0

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Divide both sides by 10:

\frac{30}{10}=a

3=a

So a=3 and b=5.

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