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Solving for ΔY

So I'm doing economics and it's been a good 10 years since I've done any algebra. I'm having difficulty rearranging equations. In the picture it shows the equation moving from:

ΔY = MPC x (ΔY-ΔT)

to

(1-MPC) x ΔY = -MPC x ΔT

to the "Final result" in the picture. I'm unsure of the steps involved to get there. I've watched many videos and googled a lot but I still can't wrap my head around it. Could someone please take me through the steps on how we arrive at the final result?

NOTE: MPC, ΔY etc are single values, not separate characters ie M, P and C

1 Answers1

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Your original equation is $$\Delta Y=MPC(\Delta Y-\Delta T)$$ Using the distributive property, you can expand the right-hand side, leading to the following result:$$\Delta Y=MPC\times\Delta Y-MPC\times\Delta T$$ Grouping all the terms that contain $\Delta Y$ gives $$\Delta Y-MPC\times\Delta Y=-MPC\times\Delta T$$ Since $\Delta Y$ is a common factor on the left-hand side, you can factorize it out as follows: $$\Delta Y(1-MPC)=-MPC\times\Delta T$$ Since you want to make $\Delta Y$ the subject of the formula, you can divide through by $(1-MPC)$ to obtain $$\Delta Y=\frac{-MPC}{1-MPC}\times\Delta T$$ and that is your answer. I hope it helps.

Poisson
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  • Thanks very much for your response, I have another question: "Since ΔY is a common factor on the left-hand side, you can factorize it out as follows: ΔY(1−MPC)=−MPC×ΔT" So this is step that stumped me, I don't understand how ΔY−MPC×ΔY can become ΔY(1−MPC). What's the rule or reasoning behind why this is? – Adam Killeen Sep 24 '20 at 06:47
  • @AdamKilleen You just factor out $\Delta Y$. Suppose you have $a\cdot b+a\cdot c$. Both summnds has the common factor $a$. Now you can factor out $a$. That means you divide every summand by $a$ and put it in front the brackets: $a\cdot b+a\cdot c=a\cdot (b+c)$. In your case $a=\Delta Y, b=1$ and $c=MPC$. If you have no further question please accept the answer $(\color{limegreen}\checkmark)$ – callculus42 Sep 24 '20 at 15:24