System of Equations - Multiplying by some constant

Question

Problem: Consider the system of equation $$\begin{cases}a+b+c=6\\a+2b+3c=14\\2a+5b+8c=36\end{cases}\tag{1}$$ Find constants $k_1,k_2$ such that $k_1$ times the first equation in the system, plus $k_2$ times the second equation of the system gives us the third equation in the system.

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From the words, I have come up with $$\begin{cases}k_1(a+b+c)+k_2(a+2b+3c)=2a+5b+8c\\6k_1+ 14k_2=36\end{cases}\tag{2}$$

But now, I'm left with more variables in a system, and I also have that feeling that I misinterpreted the problem and wrote the equation wrong.

So I'm wondering how you would write the equation, and if there is a simple and efficient way to solve the system with either substitution or elimination.

Thank you.

Answer 1

You've got:

$$k_1(a+b+c)+k_2(a+2b+3c)=2a+5b+8c$$

Or equivalently:

$$\color{red}{(k_1+k_2)}a+\color{blue}{(k_1+2k_2)}b+\color{green}{(k_1+3k_2)}c=\color{red}{2}a+\color{blue}{5}b+\color{green}{8}{c}$$

Equate coefficients

We need:

$$\color{red}{*}k_1+k_2=2$$ $$\color{blue}{*}k_1+2k_2=5$$ $$\color{green}{*}k_1+3k_2=8$$

We must also have the other equation that you mentioned satisfied:

$$6k_1+14k_2=36$$

There are $4$ equations but only two unknowns. So look at only system of 2 equations and solve that system, then check to see if it holds for the third equation and fourth.

$\color{blue}{*}$ minus $\color{red}{*}$ gives $k_2=3$. Back substitution gives $k_1=-1$. We check to see if these values satisfy all four equations, in which case we see it does.