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# Agile Algebra

If $y=x+4$ then $x+5=y+1$ and $x+3=y-1$. This means that we can write the original equation as $(y+1)^2+(y-1)^2=6$.

The "median" of these brackets is equal to $x-5\frac 1 2$, so try using a substitution of $t=x-5\frac 1 2$. You might find that a further substitution is helpful.

If $z=x+\dfrac 1 x$ then $z^2=\left(x+\dfrac {1} {x}\right)^2=x^2 + 2 + \dfrac 1 {x^2}$

You can rearrange the terms to get $x^2 +17 +\dfrac 1{x^2} - 8x - \dfrac 8 x=0$.

You can multiply the $(4x+3)$ bracket by $2$ and the $(x+1)$ bracket by $8$ - this will mean that the left hand side of the equation has multiplied by $16$, so do the same to the right hand side!

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Age 16 to 18

Challenge Level

Equations can be difficult to solve by direct attack but if you look for symmetric features and make substitutions they might become much easier to solve.

Consider the equation:$$(x+5)^2+(x+3)^2=6$$

**Expand the brackets and solve the resulting quadratic equation, leaving your answers in exact form**

The algebra here can get a little messy. Perhaps we can use a substitution to turn the equation into an easier one.

**Use the substitution $y=x+4$ to rewrite the equation in terms of $y$.**

*Why do you think that we have suggested using $y=x+4$? How does this substitution relate to the original equation?*

If $y=x+4$ then $x+5=y+1$ and $x+3=y-1$. This means that we can write the original equation as $(y+1)^2+(y-1)^2=6$.

**Expand your brackets and solve for $y$, and then for $x$.**

**Which method was easiest?**

Try using substitutions to simplify and then solve these equations:

1. $(x + 3)^4 + (x + 5) ^4 = 34$

2. $(x-1)(x-3)(x-5)(x-7)+15=0$

3. $(x-4)(x-5)(x-6)(x-7)=1680$

The "median" of these brackets is equal to $x-5\frac 1 2$, so try using a substitution of $t=x-5\frac 1 2$. You might find that a further substitution is helpful.

Now lets consider a different equation: $$x^4-8x^3 + 17x^2-8x+1=0$$

There is symmetry in the coefficients of the equation. Dividing by $x^2$ gives: $$x^2 -8x+17 -\dfrac 8 x +\dfrac 1{x^2}=0$$

Consider a substitution of $z=x+\dfrac 1 x$. What will $z^2$ be?

If $z=x+\dfrac 1 x$ then $z^2=\left(x+\dfrac {1} {x}\right)^2=x^2 + 2 + \dfrac 1 {x^2}$

Can you write $x^2 -8x+17 -\dfrac 8 x +\dfrac 1{x^2}=0$ in terms of $z$ and $z^2$?

You can rearrange the terms to get $x^2 +17 +\dfrac 1{x^2} - 8x - \dfrac 8 x=0$.

You should find that you have a quadratic in $z$ which you can solve to get two solutions. For each solutions for $z$, substitute it into $z=x+\dfrac 1 x$ and solve for $x$.

Try using substitutions to simplify and then solve these equations:

4. $x^4-2x^3 -6x^2-2x+1=0$

5. $x^4-2x^3+2x^2-2x+1=0$

**Extension**

Try using a substitution to help solve this equation. There is a hint available!

$$(8x+7)^2(4x+3)(x+1)=\frac{9}{2}$$

You can multiply the $(4x+3)$ bracket by $2$ and the $(x+1)$ bracket by $8$ - this will mean that the left hand side of the equation has multiplied by $16$, so do the same to the right hand side!

NOTES AND BACKGROUND

This is an example of a process which occurs frequently in mathematics. Let's refer to two frames of reference as A and B and say we have a problem stated in A, then the technique is to map the given relationships to B, work in B and then map the results back to A. All these equations have symmetry of one sort or another. By using the symmetry to make a substitution you can change the variable
and get an equation which is easier to solve. After that you have to use the solutions you have found and go back to find the corresponding solutions of the original equation.

To find out more about this general technique see the article "The Why and How of Substitution".