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3 years ago

Find two consecutive even integers such that four times the first plus three times the second is 20

1 answer:

8 0

2 consecutive even integers : x and x + 2

4(x) + 3(x + 2) = 20
4x + 3x + 6 = 20
7x + 6 = 20
7x = 20 - 6
7x = 14
x = 14/7
x = 2

x + 2 = 2 + 2 = 4

so ur first number (x) = 2 and ur second number (x + 2) = 4

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Answer:

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Step-by-step explanation:

This is a hard one

We have to use the rational root theorem

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We then plug them into the equation to see if any of them work

The equation isn't true when plugging 1, but is true when plugging in 1/2

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Then we divide the original equation by (2x-1) (you can use synthetic division or long division, it would be hard to type out the process for that) to get $x^2+4x+1$

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We already know the solution for the first half of the equation (1/2)

So the final answers are:

$x=\sqrt{3}-2, -\sqrt{3}-2, \frac{1}{2}$

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Step-by-step explanation:

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