National 5 Maths
Simultaneous Equations

First number the equations:

$$ \begin{eqnarray} &&2x &+& 3y &=& 4&&\:\:①\\[6pt] &&3x &-& y &=& 17&&\:\:② \end{eqnarray} $$

Multiply one or both of the equations through to create ± the same coefficient of one of the variables. In this example, it is easier to match the \(y\) terms as it only requires one of the equations to be multiplied.

$$ \begin{eqnarray} ①:\:\:&&2x &+& 3y &=& 4&&\\[6pt] 3\times②:\:\:&&9x &-& 3y &=& 51&&\:\:③ \end{eqnarray} $$

Now, because the coefficients of \(y\) have different signs, we add the two equations to eliminate \(y\):

$$ \begin{eqnarray} ①+③:\:\:&& 11x &=& 55&&\\[6pt] && \:x &=& 5&& \end{eqnarray} $$

Now we choose the easier looking of the original equations and substitute \(x=5\) into it to find \(y\):

$$ \begin{align} \small\textsf{Subs }\normalsize②:\:\:3x-y &= 17\\[6pt] 3(5)-y &= 17\\[6pt] 15-y &= 17\\[6pt] -y &= 17-15\\[6pt] -y &= 2\\[6pt] y &= -2 \end{align} $$

Video solution by Clelland Maths 

First number the equations:

In this example, we can't avoid multiplying both equations. We could do 5\(\times\)① and 3\(\times\)② to eliminate \(x\) or 3\(\times\)① and 4\(\times\)② to eliminate \(y\). It doesn't really matter which we choose. Let's eliminate \(y\):

$$ \begin{eqnarray} 3\times①:\:\:&&9x &-& 12y &=& 3 && \:\:③\\[6pt] 4\times②:\:\:&&20x &-& 12y &=& 36 && \:\:④ \end{eqnarray} $$

Now, because the coefficients of \(y\) have the same signs, we subtract the two equations to eliminate \(y\). Subtracting ③ – ④ would give two negatives, so it might be tidier to subtract upside down and do ④ – ③ instead:

$$ \begin{eqnarray} ④-③:\:\:11x &=& 33\\[6pt] x &=& 3 \end{eqnarray} $$

Now we choose the easier looking of the original equations and substitute \(x=3\) into it to find \(y\):

$$ \begin{align} \small\textsf{Subs }\normalsize①:\:\:3x-4y &= 1\\[6pt] 3(3)-4y &= 1\\[6pt] 9-4y &= 1\\[6pt] 4y &= 8\\[6pt] y &= 2 \end{align} $$

So the point of intersection of the two straight lines is \((3,2)\).

Video solution by Clelland Maths 

$$ \begin{eqnarray} &&3x &+& 2y &=& 17&&\:\:①\\[6pt] &&2x &+& 5y &=& 4&&\:\:②\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} 2\times①:\:\:&&6x &+& 4y &=& 34&&\:\:③\\[6pt] 3\times②:\:\:&&6x &+& 15y &=& 12&&\:\:④\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} ③-④:\:\:-\!11y &=& 22\\[6pt] y &=& -\!2 \end{eqnarray} $$

$$ \begin{align} \small\textsf{Subs }\normalsize①:\:\:3x+2y &= 17\\[6pt] 3x+2(-2)&= 17\\[6pt] 3x-4 &= 17\\[6pt] 3x &= 17+4\\[6pt] 3x &= 21\\[6pt] x &= 7 \end{align} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &&4x &+& 5y &=& -\!3&&\:\:①\\[6pt] &&6x &-& 2y &=& 5&&\:\:②\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} 2\times①:\:\:&&8x &+& 10y &=& -\!6&&\:\:③\\[6pt] 5\times②:\:\:&&30x &-& 10y &=& 25&&\:\:④\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} ③+④:\:\:38x &=& 19\\[6pt] x &=& \small\frac{19}{38}\\[6pt] x &=& \small\frac12 \end{eqnarray} $$

$$ \begin{align} \small\textsf{Subs }\normalsize②:\:\:6x-2y &= 5\\[6pt] 6\small\left(\frac12\right)\normalsize-2y&= 5\\[6pt] 3-2y &= 5\\[6pt] -\!2y &= 5-3\\[6pt] -\!2y &= 2\\[6pt] y &= -\!1 \end{align} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &&5c &+& 2d &=& \ 4&&\:\:①\\[6pt] &&4c &-& 3d &=& 17&&\:\:②\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} 3\times①:\:\:&&15c &+& 6d &=& 12&&\:\:③\\[6pt] 2\times②:\:\:&&\ 8c &-& 6d &=& 34&&\:\:④\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} ③+④:\:\ 23c &=& 46\\[6pt] c &=& \small\frac{46}{23}\\[6pt] c &=& 2 \end{eqnarray} $$

$$ \begin{align} \small\textsf{Subs }\normalsize②:\:\:4c-3d &= 17\\[6pt] 8-3d&= 17\\[6pt] -\!3d &= 17-8\\[6pt] -\!3d &= 9\\[6pt] d &= \small\frac{9}{-3}\\[6pt] d &= -\!3 \end{align} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &&2x &+& 3y &=& \ 8&&\:\:①\\[6pt] &&5x &+& 2y &=& -\!2&&\:\:②\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} 2\times①:\:\:&&\ 4x &+& 6y &=& 16&&\:\:③\\[6pt] 3\times②:\:\:&&15x &+& 6y &=& -\!6&&\:\:④\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} ③-④:\:\ -\!11x &=& 22\\[6pt] x &=& \small\frac{22}{-11}\\[6pt] x &=& -\!2 \end{eqnarray} $$

$$ \begin{align} \small\textsf{Subs }\normalsize①:\:\:2x+3y &= 8\\[6pt] -\!4+3y&= 8\\[6pt] 3y &= 8+4\\[6pt] 3y &= 12\\[6pt] y &= 4 \end{align} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} &&2p &-& 7r &=& 11&&\:\:①\\[6pt] &&3p &+& 2r &=& 4&&\:\:②\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} 3\times①:\:\:&&6p &-& 21r &=& 33&&\:\:③\\[6pt] 2\times②:\:\:&&6p &+& 4r &=& 8&&\:\:④\\[12pt] \end{eqnarray} $$

$$ \begin{eqnarray} ③-④:\:\:-\!25r &=& 25\\[6pt] r &=& -\!1 \end{eqnarray} $$

$$ \begin{align} \small\textsf{Subs }\normalsize①:\:\:2p-7(-1) &= 11\\[6pt] 2p+7 &= 11\\[6pt] 2p &= 4\\[6pt] p &= 2 \end{align} $$

Video solution by Clelland Maths 

Using \(a\) for the price in £ of an adult ticket and \(c\) for the price in £ of a child ticket, we have:

So an adult ticket costs £7.50 and a child ticket costs £4.25.

Video solution by Clelland Maths 

(a)  We need to choose our letters. Let's use \(c\) for the amount of material in a cloak and \(d\) for the amount of material in a dress.

So we get the equation: \(2c+3d=9.6\)

(b)  The second equation is \(3c+4d=13.3\)

(c)  Now we need to solve the two equations simultaneously.

$$ \begin{eqnarray} \small\textsf{}\normalsize\:\:\: && 2c+3d &=& 9.6&&\:\:①\\[6pt] \small\textsf{}\normalsize\:\:\: && 3c+4d &=& 13.3&&\:\:②\\[12pt] 3\times①:\:\:\:\: && 6c+9d &=& 28.8 && \:\:③\\[6pt] 2\times②:\:\:\:\: && 6c+8d &=& 26.6 && \:\:④\\[18pt] \end{eqnarray} $$

$$ \begin{align} ③-④:\:\:\:\:\:\:d &= 2.2\\[12pt] \small\textsf{Subs }\normalsize①:\:\:\:\:2c+3(2.2) &= 9.6\\[6pt] 2c+6.6 &= 9.6\\[6pt] 2c &= 9.6-6.6\\[6pt] 2c &= 3\\[6pt] c &= 1.5 \end{align} $$

So a cloak uses 1.5 square metres of material and a dress uses 2.2 square metres of material.

Video solution by Clelland Maths 

(a)  We need to choose our letters. Let's use \(c\) for the weight of one bag of cement and \(g\) for the weight of one bag of gravel.

So we get the equation: \(7c+3g=215\)

(b)  The second equation is \(5c+4g=200\)

(c)  Now we need to solve the two equations simultaneously.

$$ \begin{eqnarray} \small\textsf{}\normalsize\:\:\: && 7c+3g &=& 215&&\:\:①\\[6pt] \small\textsf{}\normalsize\:\:\: && 5c+4g &=& 200&&\:\:②\\[12pt] 4\times①:\:\:\:\: && 28c+12g &=& 860 && \:\:③\\[6pt] 3\times②:\:\:\:\: && 15c+12g &=& 600 && \:\:④\\[24pt] \end{eqnarray} $$

$$ \begin{align} ③-④:\:\:\:\:\:\:13c &= 260\\[6pt] c &= \small\frac{260}{13}\\[6pt] c &= 20\\[18pt] \small\textsf{Subs }\normalsize①:\:\:\:\:7(20)+3g &= 215\\[6pt] 140+3g &= 215\\[6pt] 3g &= 215-140\\[6pt] 3g &= 75\\[6pt] g &= \small\frac{75}{3}\\[6pt] g &= 25\\[6pt] \end{align} $$

So one bag of cement weighs 20 kg and one bag of gravel weighs 25 kg.

Video solution by Clelland Maths 

(a)  We need to choose our letters. Let's use \(m\) for the price of one mango and \(a\) for the price of one apple. We can work in pounds or pence in this question, but to avoid decimals, perhaps it's easier to work in pence.

So we get the equation: \(4m+3a=425\)

(b)  The second equation is \(5m+2a=470\)

(c)  Now we need to solve the two equations simultaneously.

$$ \begin{eqnarray} \small\textsf{}\normalsize\:\:\: && 4m+3a &=& 425&&\:\:①\\[6pt] \small\textsf{}\normalsize\:\:\: && 5m+2a &=& 470&&\:\:②\\[12pt] 2\times①:\:\:\:\: && 8m+6a &=& 850 && \:\:③\\[6pt] 3\times②:\:\:\:\: && 15m+6a &=& 1410 && \:\:④\\[24pt] \end{eqnarray} $$

$$ \begin{align} ④-③:\:\:\:\:\:\:7m &= 560\\[6pt] m &= \small\frac{560}{7}\\[6pt] m &= 80\\[18pt] \small\textsf{Subs }\normalsize①:\:\:\:\:4(80)+3a &= 425\\[6pt] 320+3a &= 425\\[6pt] 3a &= 425-320\\[6pt] 3a &= 105\\[6pt] a &= \small\frac{105}{3}\\[6pt] a &= 35\\[6pt] \end{align} $$

So a mango costs 80p and an apple costs 35p.

Video solution by Clelland Maths