Could anyone help me solve this simultaneous equation: $$6x + 7y = 12.5\tag1$$ $$7x + 5y = 14\tag2$$ I have watched loads of videos on YouTube, and still don't understand how to do it. I am trying to learn it. And practice with different equations. But don't understand how to do it. So any help will be much appreciated.

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    $\begingroup$ Now take $7(2)-5(1)$! $\endgroup$ – Parcly Taxel May 25 '17 at 13:03
  • $\begingroup$ What about it is confusing you? Give a bit more context so we can answer with something you can understand. $\endgroup$ – Sean Roberson May 25 '17 at 13:04

We have \begin{align}6x + 7y &= 12.5\tag{1}\\ 7x + 5y &= 14\tag{2}\end{align}

One way to solve this, would be to rearrange equation $(1)$ for $x$:

\begin{align}6x + 7y &= 12.5\\ 6x&=12.5-7y\\ x&=\frac{12.5-7y}{6}\end{align}

We can then put this into equation $(2)$ and solve for $y$:

\begin{align}7x + 5y &= 14\\ 7\left(\frac{12.5-7y}{6}\right)+5y&=14\\ \frac{87.5-49y}{6}+5y&=14\\ 87.5-49y+30y&=84\\ 87.5-19y&=84\\ 19y&=3.5\\ y&=\frac{3.5}{19}\\ &=\frac{7}{38}\end{align}

Finally we can put this into equation $(1)$ to find that \begin{align}6x + 7\left(\frac{7}{38}\right) &= 12.5\\ 6x+\frac{49}{38}&=12.5\\ 6x&=12.5-\frac{49}{38}\\ 6x&=\frac{213}{19}\\ x&=\frac{71}{38}\end{align}

We can then check this using equation $(2)$:

\begin{align}7x + 5y &=7\left(\frac{71}{38}\right)+5\left(\frac{7}{38}\right)\\ &=\frac{497}{38}+\frac{35}{38}\\ &=\frac{532}{38}\\ &=14\end{align}

Therefore we can be confident that the answer is $x=\dfrac{71}{38}$ and $y=\dfrac{7}{38}$

  • $\begingroup$ Thank you for this. I am going to keep going through this until I fully understand it. And then try it on another equation. I really appreciate it. $\endgroup$ – Liam May 25 '17 at 13:30
  • $\begingroup$ @Liam No worries. Make use of this site and ask other questions if you get stuck again. Also, in case you are not aware, you can upvote answers you like, and can also mark them as 'acecepted' (basically means it is the best answer for your question in your eyes) by clicking the checkmark next to the answer. This is how the site makes sure the best answers are seen first, when other people seach for a problem $\endgroup$ – lioness99a May 25 '17 at 13:33
  • $\begingroup$ I have marked this as accepted. I just wanted to ask one more question. I am seeing an answer in the comments that is not a fraction. Which confuses me a bit. Is it be ok to solve an equation like this with a fraction? And is this the best way to solve this equation? Thanks $\endgroup$ – Liam May 25 '17 at 13:58
  • $\begingroup$ @Liam It is always best to write an answer as a fraction, if this is the most accurate answer. For example, $\frac{71}{38}$ is in its simplest form. We could write it as $1.868421052631578947368421052631578947368421052631578947368\ldots$ but this would be less accurate. The only exception to this rule is if the question asked for the answer to be given to a specific number of significant figures. Even in this case, you would work with the fractions right up until the moment you wrote down the answer, when you would round anything that wasn't exact $\endgroup$ – lioness99a May 25 '17 at 14:03
  • $\begingroup$ Hello, thank you for your help. Very much appreciated. $\endgroup$ – Liam May 25 '17 at 14:50

A method that always works for such (linear) equations is to isolate one of the variables in one of the equation, say $y$ in $(1)$, and then insert that expression for $y,$ which is now written in terms of $x,$ into the other equation. Then you will have an equation with only one variable, $x,$ which you can then solve.

Having done this, you can insert that value for $x$ into the expression for $y,$ where you had isolated $y.$ Now you have both the $x$ and $y$.

Try it out, and if it is still causing you trouble, leave me a comment below and I'll show the full solution.

Good luck!

A side-note: Often, however, there are simpler ways of doing things, as @ParclyTaxel points out in the comments above. What is being referred to is multiplying equation $(2)$ with seven and then subtracting five times equation $(1).$ This is done to achieve a common factor, allowing the cancellation of a variable, as such:

\begin{align} 7(2):\quad & 49x+35y=7\cdot 14\\ 5(1):\quad & 30x+35y=5\cdot 12.5 \end{align} Subtracting then gives $$49x+35y-30x-35y=7\cdot 14-5\cdot 12.5$$ simplifying to $$19x=35.5.$$

Shortcuts like this comes with experience. The way described above, however, is straight-forward application of an algorithm, and doesn't require any "ideas", so it always works.

  • $\begingroup$ This is the way I was trying to do it from watching YouTube. But could not get an answer. I will try again this way and try to get it correct. Thank you. $\endgroup$ – Liam May 25 '17 at 13:28

Here is an another example, then you can apply it to your problem:

(i) $x+y=4$

(ii) $2x+y=7$

Solving (i) for $x$ gives us $x=4-y$. Now put this into (ii) and you get $2(4-y)+y=7$ and you can now easily solve it for $y$. You should get $y=1$ and finally $x=4-1=3$.


we can also write $$x+\frac{7}{6}=\frac{25}{12}$$ $$x+\frac{5}{7}y=2$$ multiplying by $$-1$$ we have $$-x-\frac{7}{3}y=-\frac{25}{12}$$ $$x+\frac{5}{7}y=2$$ adding both we obatain: $$\frac{5}{7}y-\frac{7}{3}y=2-\frac{25}{12}$$ can you finish? simplifying the last equation we obtain: $$-\frac{34}{21}y=-\frac{1}{12}$$ $$y=\frac{7}{38}$$ and then you can compute $$x$$

  • $\begingroup$ I can not finish this. I have tried but do not understand it. Thank you for commenting though. $\endgroup$ – Liam May 25 '17 at 13:29
  • $\begingroup$ oh sorry, what Kind of questions do you have? $\endgroup$ – Dr. Sonnhard Graubner May 25 '17 at 13:35
  • $\begingroup$ I am getting a bit confused because there is 3 different ways I am seeing from the comments of solving this equation. And different answers. So I am just trying to understand them. Could you finish yours please? So I can try and understand it. Thank you. $\endgroup$ – Liam May 25 '17 at 13:49

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