Question #15aab

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Question #15aab

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Answer 1 · 2016-11-21T23:41:23

$p = 3 and q = 4$ $p=3" and " q=4$

Explanation:

The question to be solved is.

$5 p - 7 q = - 13 \to (1)$ $5p-7q=-13to(1)$
$2 p + 7 q = 34 \to (2)$ $2p+7q=34to(2)$

The aim here is to eliminate one of the variables from the equations by performing operations on them. We then have a one variable equation which we can solve. Then substitute this value into either one of the 2 given equations and solve for the other variable.

I have labelled the equations to assist.

Note that in (1) there is a - 7q and in (2) there is + 7q

Adding these together will eliminate the variable q, which is what we want.

Add (1) and (2) together term by term on both sides.

$\Rightarrow 7 p + 0 = 21 \Rightarrow 7 p = 21 \Rightarrow p = 3$ $rArr7p+0=21rArr7p=21rArrp=3$

Now substitute p = 3 into (1) or (2). I'll choose (2) you can perhaps try (1).

$\Rightarrow (2 \times 3) + 7 q = 34$ $rArr(2xx3)+7q=34$

$\Rightarrow 6 + 7 q = 34 \Rightarrow 7 q = 28 \Rightarrow q = 4$ $rArr6+7q=34rArr7q=28rArrq=4$

Hope this helps.

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Question #15aab

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