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I'm doing an 8th grade math text book, and came across this problem:

$4a + $ $ 5b + $ $9c$ = 36; and 7a + 9b + 17c = 66, what is a + b + c = ?

The first thing I could notice was that, " How could you find a suitable solution for two equations with THREE variables " But still hoped that there must be some way to solve it.

If I've forgotten any basic concept, please let me know.

Thank you.

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$4a+ 5b +9c = 36$;
$7a + 9b + 17c = 66$;

$\implies$ (multiply $1$st equation by $2$)

$8a+ 10b +18c = 72$;
$7a + 9b + 17c = 66$;

Then subtract equations:

$a+b+c= (8a+10b+18c) - (7a+9b+17c) = ...$ .

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  • $\begingroup$ Thank you very much, this helped. But why do you need to multiply by two only? or is it because 2 is SUITABLE for THIS problem? please explain $\endgroup$ – Ramana Apr 15 '14 at 6:40
  • $\begingroup$ Yes, it is suitable for THIS problem. Here you need to note that coefficients are almost proportional. $\endgroup$ – Oleg567 Apr 15 '14 at 6:43
  • $\begingroup$ I got it right. Thankyou again. $\endgroup$ – Ramana Apr 15 '14 at 6:45
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    $\begingroup$ If you have $2$ equations with $3$ variables, and you need to write something like $a+b+c$, in most cases it must be something like $K_1*LHS_1+K_2*LHS_2$. Here $K_1=2, K_2=-1$. (LHS - "left-hand side of equation") $\endgroup$ – Oleg567 Apr 15 '14 at 6:46
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    $\begingroup$ You are welcome :) I am happy that so young person is user of MathStackexchange. $\endgroup$ – Oleg567 Apr 15 '14 at 6:47
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Use the two equations and solve them for $b$ and $c$ considering that $a$ is a constant (these are two linear equations for two unknowns). You should arrive to $$b=\frac{9}{2}-\frac{5 a}{4}$$ $$c=\frac{a}{4}+\frac{3}{2}$$ I am sure that you can take from her.

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$$36x+66y=x(4a+5b+9c)+y(7a+9b+17c)$$ $$36x+66y=a(4x+7y)+b(5x+9y)+c(9x+17y)\ \ \ \ (1)$$

So we need to choose $x,y$ such that $$4x+7y=5x+9y=9x+17y=1$$

Solving $\displaystyle 4x+7y=1\ \ \ \ (2),5x+9y=1\ \ \ \ (3)$ for $x,y$ we get $x=2,y=-1$ which satisfies $9x+17y=1\ \ \ \ (4)$

Set the values of $x,y$ in $(1)$

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  • $\begingroup$ For what are the x, and y there $\endgroup$ – Ramana Apr 15 '14 at 6:23
  • $\begingroup$ @Ramana, Please find the edited version $\endgroup$ – lab bhattacharjee Apr 15 '14 at 6:28
  • $\begingroup$ I cannot find it :( help $\endgroup$ – Ramana Apr 15 '14 at 6:29
  • $\begingroup$ @Ramana, Please pinpoint your area of confusion $\endgroup$ – lab bhattacharjee Apr 15 '14 at 6:31
  • $\begingroup$ Why should we bring in the variables x,y here to solve these two equations? $\endgroup$ – Ramana Apr 15 '14 at 6:36

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