Equation system with two unknown variables $xy=6930$ and $y/x=1.41$

Question

I am sorry if this is too easy for you, but it's been very many decades since I had math at school. ;-) Please also consider that English is a foreign language for me.

I need to resolve X and Y for this equation system:

\begin{align} x\cdot y &= 6930\\ \frac yx &= 1.41 \end{align}

How can this be done and which are the steps to the solution? Please write in easy to understand arithmetic language (no advanced mathematical symbols please).

Many thanks in advance!

Answer 1

The first equation $$ x \cdot y = 6930 \quad (*) $$ implies that both $x$ and $y$ can not be zero.

So the second equation $$ \frac{y}{x} = 1.41 \quad (**) $$ is equivalent (has the same solutions) to $$ y = 1.41 \cdot x $$ We can insert this into the first equation and get $$ x \cdot (1.41 \cdot x) = 6930 $$ which simplifies to $$ x^2 = \frac{6930}{1.41} $$ or $$ x = \pm \sqrt{\frac{6930}{1.41}} $$ where the $\pm$ is short hand notation for two solutions, one positive, one negative.

Finally we get $$ y = 1.41 \cdot x = 1.41 \cdot \pm \sqrt{\frac{6930}{1.41}} = \pm \sqrt{1.41 \cdot 6930} $$ So we got four solutions in total, however equation $(*)$ or $(**)$ will only permit pairs of same signs, as we need a positive product or fraction, so we end up with two solutions: $$ (x, y) = \pm \left( \sqrt{\frac{6930}{1.41}}, \sqrt{1.41 \cdot 6930}\right) $$

Update:

Here you can fiddle with GeoGebra: link

This free software allows you to solve the problem graphically, symbolic and numeric.

Answer 2

If $y/x = 1.41$, then $y = 1.41x$. So plugging this in the first equation yields $1.41x^2 = 6930$ and thus $x = \pm \sqrt{6930/1.41}$. Now it should be easy to find also $y$.

Answer 3

First divide the first equation by X:

$$Y = 6930/X$$

Then substitute the Y into the 2nd equation.

$$(6930/X)/X=1.41$$

Multiply by X:

$$6930/X = 1.41X$$

... again

$$6930=1.41X^2$$

Divide by 1.41

$$4914.89=X^2$$

Take square root:

$$X= +/-70.10$$

Solve the Y by yourself :).

Answer 4

Start with$$\begin{align*} & xy=6930\tag1\\ & \tfrac yx=1.41\tag2\end{align*}$$ Multiplying the numerator and denominator of $(2)$ by $y$ gives$$\begin{align*} & \dfrac {y^2}{xy}=1.41\tag3\\ & y^2=1.41\cdot xy\tag4\\ & y^2=1.41\cdot 6930\tag5\\ & y=\pm\sqrt{9771.3}\end{align*}$$ Substitute that back into $(1)$ to get the corresponding $x$ values.