How do I solve these simultaneous equations using an algebraic method? \begin{align} 3x + 6y & = 30\\ x + 6y & = 20 \end{align}

It's on a old test paper and I'm kind of doing revision...

I was thinking of doing $30 - 20 = 10$. I don't know what to do after? We haven't done simultaneous questions yet in class but I'm trying to work it out myself. Can you give any hints or a formula?

Could I do trial and error? But I don't want to do that, I want to know the method.

  • $\begingroup$ x = 5 y = 2.5 Ok I just need to plot it into the equation $\endgroup$ – user61406 May 11 '13 at 17:58

$$3x + 6y = 30\tag{1}$$

$$\;x + 6y = 20\tag{2}$$

Subtracting term for term $(2)$ from $(1)$ gives us $$(3x + 6y) - (x + 6y) = 30 - 20\tag{3}$$ $$3x - x + 6y - 6y = 10,\; $$ (Very nice, since we eliminate $y$!) Now we can solve the resulting equation for $x$

$$2x + 0\cdot y = 10 \implies x = 5$$

Now just solve for $y$, using, say, equation $(2)$: $$x + 6y = 20\;\;\text{and} \;\;x = 5 \implies 5 + 6y = 20$$ $$ \implies 6y = 20 - 5 \implies 6y = 15 \implies y = \frac {15}{6} = \frac 52$$

$$x = 5 \quad \text{and }\quad y = \frac 52$$

  • $\begingroup$ +1 for doing the subtraction of both sides properly rather than magically. $\endgroup$ – JP McCarthy Jun 24 '13 at 14:47

Subtract the two equations. We then get $$\color{red}{(3x+6y)} - \color{blue}{(x+6y)} = \color{red}{30} - \color{blue}{20} \implies (3x-x) + (6y-6y) = 10 \implies 2x = 10 \implies x = 5$$ Now plug in $x=5$ in one of the equations, say in the equation $\color{blue}{(x+6y) = 20}$. We then get that $$5+6y = 20 \implies 6y = 15 \implies y = \dfrac{5 \times3}{2 \times 2} = \dfrac52$$


Note that every coefficient in the first equation is a multiple of $3$, so you can divide through by $3$ and replace $3x+6y=30$ with the equivalent equation $x+2y=10$. You now have the system

$$\left\{\begin{align*} &x+2y=10\\ &x+6y=20\;. \end{align*}\right.$$

Now it’s very apparent that if you subtract the top equation from the bottom one, you’ll get an equation that does not contain $x$:


or, after simplification,


You can easily solve that for $y$ and then substitute the result back into either of the original equations and solve the resulting equation for $x$.

Even if the coefficients in the first equation had not all been multiples of $3$, I could have used this approach. Suppose that the first equation had been $3x+5y=32$; dividing through by $3$ would have left me with the system

$$\left\{\begin{align*} &x+\frac53y=\frac{32}3\\\\ &x+6y=20\;, \end{align*}\right.$$

and I can still subtract one equation from the other to get a single equation without $x$ in it. The arithmetic is a little messier, but the principle is exactly the same and can be applied to any system of this kind.


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