Solution

$a)$ This is a nonlinear equation as $x$ is squared (raised to the power of $2$).

$b)$ This is a linear equation as it is in the correct form: neither $x \text { nor } y$ is raised to a power other than one or zero and their coefficients are constants.

$c)$ This is a nonlinear equation as $x \text{ and } y$ are multiplied together: we have a product of variables.

Solution

The aim is to collect the $x$ terms on one side and the constants on the other. Begin by multiplying through by $3$ to get rid of the fraction. \begin{align} 3 \times (4x-1) &= 3 \times \frac{1}{3}(2+2x) \\ 12x - 3 &= 2+2x \end{align}

Subtract $2x$ from both sides so that there is no $x$ term on the right. \begin{align} 12x - 3 - 2x &= 2 + 2x - 2x \\ 10x - 3 &= 2 \\ \end{align}

Add $3$ to both sides so that the $x$ term is on its own on the left. \begin{align} 10x - 3 + 3 &= 2 + 3 \\ 10x &= 5 \end{align}

Finally, divide by $10$ to get an expression for $x$. \begin{align} \frac{10x}{10} &= \frac{5}{10} \\ \\ x &= \frac{1}{2} \end{align}

Solution

Take the second fraction over to the right hand side and then cross multiply. \begin{align} \frac{3}{3-a} &= -\frac{4}{2a+1} \\ 3(2a+1) &= -4(3-a) \\ 6a+3 &=-12+4a \end{align} Subtract $4a$ from both sides. \begin{align} 6a + 3 - 4a &= -12 + 4a - 4a \\ 2a + 3 &= -12 \end{align}

Subtract $3$ from both sides. \begin{align} 2a + 3 - 3 &= -12 - 3 \\ 2a &= -15 \end{align}

Divide by $2$ to get a solution for $a$. \begin{align} \frac{2a}{2} &= - \frac{15}{2} \\ \\ a &= - \frac{15}{2} \end{align}

Solution

a) Substituting $p=10$ into the supply equation gives: \begin{align} q^S&=p+10\\ &=10+10\\ &=20 \end{align}

So $20$ kilos of bananas would be supplied at a price of $£10$.

b) We must first rearrange the supply equation to make $p$ the subject: \begin{align} q^S&=p+10\\ \Rightarrow p&=q^S-10 \end{align}

Substituting $q^S=30$ into this rearranged equation gives: \begin{align} p&=30-10\\ &=20 \end{align} so if $30$ kilos are supplied, the price must be $£20$.

Solution

a) The budget constraint can be expressed as a linear equation: \[2.50=0.15M+0.1G\] where $£2.50$ is the total amount of money (in $£$) Joe has to spend and $M$ and $G$ are the amount of sugar mice and bags of Gumdrops respectively that Joe buys. Rearranging the budget constraint to make $G$ the subject gives: \begin{align} 0.1G&=2.50-0.15M\\ \Rightarrow G&=\frac{2.50}{0.1}-\frac{0.15M}{0.1}\\ &=25-1.5M \end{align}

We can now plot the (rearranged) budget constraint $G=25-1.5M$:

b) We can find out how many bags of Gumdrops Joe can have if he buys $2$ sugar mice by substituting $M=2$ into the rearranged budget constraint: \begin{align} G&=25-1.5\times 2\\ &=22 \end{align}

So for $£2.50$ Joe can have $2$ sugar mice and $22$ bags of Gumdrops.

c) The budget constraint with the new price is: \[2.50=0.2M+0.1G\] Rearranging this budget constraint to make $G$ the subject gives: \begin{align} 2.50&=0.2M+0.1G\\ \Rightarrow G&=\frac{2.50}{0.1}-\frac{0.2M}{0.1}\\ &=25-2M \end{align}

We can now plot the (rearranged) budget constraint $G=25-2M$:

The new budget constraint is shown by the blue line. We can see that the higher price of sugar mice has made the budget constraint steeper: Joe must now give up more Gumdrops for each additional sugar mouse he wants to buy with his pocket money.

To find out how many bags of Gumdrops can Joe now have if he buys $2$ sugar mice we need to substitute $M=2$ into the new (rearranged) budget constraint. This gives: \begin{align} G&=25-2\times 2\\ &=21 \end{align}

So now Joe can have $2$ sugar mice and $21$ bags of Gumdrops. That is, for $£2.50$ Joe can have $1$ less bag of Gumdrops and the same number of sugar mice at the new price of sugar mice. This corresponds to our observations from the graph.

d) With his new amount of pocket money, Joe's budget constraint is given by: \[3=0.15M+0.1G\] Rearranging this budget constraint to make $G$ the subject gives: \[G=30-1.5M\]

We can now plot this rearranged budget constraint:

The new budget constraint is shown by the green line. We can see that the higher amount of pocket money has shifted Joe's budget constraint outwards (to the right): Joe can now have more sugar mice and more Gumdrops than he could with the amount of money and prices in part a).

To find out how many bags of Gumdrops can Joe now have if he buys $2$ sugar mice we need to substitute $M=2$ into the new (rearranged) budget constraint. This gives: \begin{align} G&=30-0.15\times 2\\ &=27 \end{align}

So now Joe can have $2$ sugar mice and $27$ bags of Gumdrops. For $£3$ Joe can have $5$ more bag of Gumdrops and the same number of sugar mice than with the amount of pocket money he usually receives and the prices from part a). This corresponds to our observations from the graph.