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This should be something I remember doing, but I can't.

What's given is $$ .15 = .7(.1)+.4a + .2b $$ and that $$.7 + a + b = 1$$

The answers on the solution sheet is $a = .1$, $ b = .2$

Can someone walk me through this and explain how to get a and b? thanks

edit: you guys were right. it should have been .4a and .2b

Kevin R.
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    I think there is a typo in the first equation. – RK01 Jul 08 '15 at 18:54
  • there appears to be an error in your problem (as it stands, the values you give are not solutions to your linear system). in the first equation did you possibly mean 7(.1) ? I think that fixes it. – lulu Jul 08 '15 at 18:57
  • Also, .15 should be 1.5 – RK01 Jul 08 '15 at 18:58
  • just saying you can't remember is not good enough. What are your thoughts so far? What have you tried? – Charlie Parker Jul 08 '15 at 19:00
  • Don't think there's a typo. This is straight off the solution sheet. What he did was , a = .3-b -> .8 = .4(3-b) + .2b -> a = .1, b = .2 – Kevin R. Jul 08 '15 at 19:05

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If $0.7+a+b=1$ then $a = 1 - b-0.7 = 0.3-b$. Therefore you can put $0.3-b$ in place of $a$.

\begin{align} 0.15 & = 0.07 +0.4a+ 0.2b \\[10pt] 0.15 & = 0.07 +0.4(\cdots)+0.2b \\[10pt] 0.15 & = 0.07 +0.4(0.3-b)+0.2b \end{align}

The last equation has only one unknown quantity. After you find that, then you can use $a=0.3-b$ to find $a$.

Michael Hardy
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