Review question

# Can we solve these equations for $a, b, x$ and $y$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6871

## Solution

Find all values of $a$, $b$, $x$ and $y$ that satisfy \begin{align*} a+b &= 1 \\ ax+by &= \frac{1}{3} \\ ax^2+by^2 &= \frac{1}{5} \\ ax^3+by^3 &= \frac{1}{7}. \end{align*}

[Hint: you may wish to start by multiplying the second equation by $x+y$.]

Let’s do what the hint suggests, and multiply the second equation by $x+y$, which yields $(ax+by)(x+y) = \frac{1}{3}(x+y),$ that is, $ax^2+by^2+(a+b)xy = \frac{1}{3}(x+y).$

Substituting in the information from the first and third equations yields $\frac{1}{5}+xy=\frac{1}{3}(x+y),$ or, rearranging, $x+y-3xy-\frac{3}{5}=0.$

Now let’s try the same trick with the third equation; we have $(ax^2+by^2)(x+y) = \frac{1}{5}(x+y),$ that is, $ax^3+by^3+xy(ax+by) = \frac{1}{5}(x+y).$

Substituting in the information from the second and fourth equations, we find that $\frac{1}{7}+\frac{xy}{3}=\frac{1}{5}(x+y),$ or, rearranging, $x+y-\frac{5}{3}xy-\frac{5}{7}=0.$

Now let $xy = z$. This gives us

$$$x+y-3z-\frac{3}{5}=0, \label{eq:1}$$$ and $$$x+y-\frac{5}{3}z-\frac{5}{7}=0. \label{eq:2}$$$

so subtracting these gives us $-\dfrac{4z}{3}+ \dfrac{4}{35}=0,$ which means that $z = \dfrac{3}{35}$.

So $y = \dfrac{3}{35x}$, and thus from (1), $x+\dfrac{3}{35x} - 3\dfrac{3}{35}-\dfrac{3}{5}=0.$ This is the quadratic equation $35x^2-30x+3=0.$ Since $x$ and $y$ are interchangeable, we take the solution to be \begin{align*} x &= \frac{30+4\sqrt{30}}{70}=\frac{3}{7}+\frac{2\sqrt{30}}{35},\\ y &= \frac{30-4\sqrt{30}}{70}=\frac{3}{7}-\frac{2\sqrt{30}}{35}. \end{align*}

We now wish to find the values of $a$ and $b$ for these values of $x$ and $y$. The second simultaneous equation with the above values of $x$ and $y$ looks like $\frac{3a}{7}+\frac{2\sqrt{30}a}{35}+\frac{3b}{7}-\frac{2\sqrt{30}b}{35}=\frac{1}{3},$ and substituting in $b=1-a$ from the first of the simultaneous equations yields $\frac{4\sqrt{30}a}{35}+\frac{3}{7}-\frac{2\sqrt{30}}{35}=\frac{1}{3}.$ Rearranging, we find that $a=\frac{1}{2}-\frac{5}{6\sqrt{30}}=\frac{1}{2}-\frac{\sqrt{30}}{36},$ and therefore $b=1-a=1-\frac{1}{2}+\frac{\sqrt{30}}{36}=\frac{1}{2}+\frac{\sqrt{30}}{36}.$ Notice that if we’d chosen $x$ and $y$ to be the other way around then $a$ and $b$ would also have been swapped.

How do we know this solution works without substituting our values in to all 4 equations to check?

How would our answer change if we were given the additional constraint that $ax^4+by^4 = \dfrac{1}{9}$?

This GeoGebra file helps us to see what is happening.

We are plotting here the three curves $ax+(1-a)y = \dfrac{1}{3}, ax^2+(1-a)y^2 = \dfrac{1}{5}$ and $ax^3+(1-a)y^3 = \dfrac{1}{7}$.

As we vary $a$, can we get the three curves to all pass through the same point?