Algebra Manipulations — Mathslive.ie

Section 1 of 5

Linear — x on one side

The job: make $x$ the subject. Whatever is on the $x$, undo it — in reverse.

(i) Warm-up with numbers

Solve $2x + 1 = 3$.

Take $1$ from both sides: $2x = 2$

Divide both sides by $2$: $x = 1$

We undid the $+1$ first, then the $\times 2$. Reverse order.

(ii) Same idea, letters instead

Make $x$ the subject of $y = 2x + 1$.

$y - 1 = 2x$

$\dfrac{y - 1}{2} = x$

$x = \dfrac{y - 1}{2}$

(iii) General form

In general, given $ax + b = c$:

$ax = c - b$

$x = \dfrac{c - b}{a}$

Must learn

1.Undo the operations in reverse order.

2.Undo $+/-$ first, then undo $\times/\div$.

YOU TRY · 1

If $y = 3x - 4$, write $x$ in terms of $y$.

Undo the $-4$ first, then the $\times 3$.

$y + 4 = 3x$

$x = \dfrac{y + 4}{3}$

Section 2 of 5

Linear — x on both sides

Make $x$ the subject of $a(x+1) = b(x+3)$.

$x$ appears on both sides. We need to gather them.

(i) Step by step

Expand both sides:

$ax + a = bx + 3b$

Gather $x$ terms on the LEFT, everything else on the RIGHT:

$ax - bx = 3b - a$

Factor $x$ out of the left side:

$x(a - b) = 3b - a$

Divide both sides by $(a - b)$:

$x = \dfrac{3b - a}{a - b}$

(ii) The pattern

After expanding, what's the trick that makes it solvable?

Pull $x$ out as a common factor: $x(a - b)$.

Then $(a-b)$ is just a number — divide both sides by it.

Must learn

1.Expand all brackets.

2.Gather all $x$ terms on ONE side.

3.Factor $x$ out.

4.Divide.

YOU TRY · 2

If $a(x - 2) = b(x + 4)$, write $x$ in terms of $a$ and $b$.

Expand → gather → factor → divide.

$ax - 2a = bx + 4b$

$ax - bx = 4b + 2a$

$x(a - b) = 4b + 2a$

$x = \dfrac{4b + 2a}{a - b}$

Section 3 of 5

Square root — kill it first

Make $x$ the subject of $t = \sqrt{\dfrac{ax + b}{x + c}}$.

There's a square root. What's always the first move?

Square both sides:

$t^2 = \dfrac{ax + b}{x + c}$

Now clear the fraction — multiply both sides by $(x + c)$:

$t^2 (x + c) = ax + b$

Expand:

$t^2 x + t^2 c = ax + b$

From here it's a linear equation with $x$ on both sides — same drill as Section 2.

Gather $x$ terms on the LEFT:

$t^2 x - ax = b - t^2 c$

Factor $x$ out:

$x(t^2 - a) = b - t^2 c$

Divide:

$x = \dfrac{b - t^2 c}{t^2 - a}$

Must learn

1.Square first — that kills the root.

2.Clear any fraction by multiplying.

3.Then it's linear — gather, factor, divide.

YOU TRY · 3

If $t = \sqrt{\dfrac{x + 1}{x - 2}}$, write $x$ in terms of $t$.

Square. Clear the fraction. Then gather $x$ terms.

$t^2 = \dfrac{x + 1}{x - 2}$

$t^2 (x - 2) = x + 1$

$t^2 x - 2t^2 = x + 1$

$t^2 x - x = 1 + 2t^2$

$x(t^2 - 1) = 1 + 2t^2$

$x = \dfrac{1 + 2t^2}{t^2 - 1}$

Section 4 of 5

Complete the square

Make $x$ the subject of $x^2 + 6x + 1 = t$.

(i) First instinct — and why it fails

Try the Section 1 method — isolate the $x$ terms, then factor:

$x^2 + 6x = t - 1$

$x(x + 6) = t - 1$

$x = \dfrac{t - 1}{x + 6}$ ($x$ still on both sides — dead end)

Factoring doesn't get us out. We need a new tool.

(ii) Squares to know cold

Quick warm-up — expand each one in your head, then tap to check.

$(x + 2)^2 = \ ?$

$(x + 2)^2 = x^2 + 4x + 4$

$(x + 3)^2 = \ ?$

$(x + 3)^2 = x^2 + 6x + 9$

$(x + 5)^2 = \ ?$

$(x + 5)^2 = x^2 + 10x + 25$

$(x - 8)^2 = \ ?$

$(x - 8)^2 = x^2 - 16x + 64$

Pattern: $(x + a)^2 = x^2 + 2ax + a^2$. The middle coefficient is $2a$, the constant is $a^2$.

(iii) The method — complete the square

Back to $x^2 + 6x + 1 = t$.

Halve the middle coefficient: $\dfrac{6}{2} = 3$. Square it: $3^2 = 9$.

Add AND subtract $9$ on the left so we change nothing:

$x^2 + 6x + 9 + 1 - 9 = t$

The first three terms now form a perfect square:

$(x + 3)^2 - 8 = t$

$(x + 3)^2 = t + 8$

$x + 3 = \sqrt{t + 8}$

$x = \sqrt{t + 8} - 3$

(iv) Another worked example

Given $y = x^2 - 10x + 3$, write $x$ in terms of $y$.

$y = x^2 - 10x + 3$

Half of $-10$ is $-5$. Square it: $25$.

$y = x^2 - 10x + 25 + 3 - 25$

$y = (x - 5)^2 - 22$

$y + 22 = (x - 5)^2$

$\sqrt{y + 22} = x - 5$

$x = \sqrt{y + 22} + 5$

Must learn

1.Quadratic in $x$? Complete the square.

2.Take HALF the coefficient of $x$. Square it.

3.Add AND subtract that number — you've changed nothing.

4.The first three terms now form $(x + a)^2$.

5.Isolate the square, then take the root.

YOU TRY · 4

Given $x^2 - 12x - 1 = y$, write $x$ in terms of $y$.

Half of $-12$ is $-6$. Square it: $36$. Add and subtract.

$x^2 - 12x + 36 - 1 - 36 = y$

$(x - 6)^2 - 37 = y$

$(x - 6)^2 = y + 37$

$x - 6 = \sqrt{y + 37}$

$x = \sqrt{y + 37} + 6$

Section 5 of 5

Substitution

Given $x = 2a$ and $x = 5b$, write $a$ in terms of $b$.

Both expressions equal $x$ — so they must equal each other.

$2a = 5b$

$a = \dfrac{5}{2}b$

That's it. This trick — substitution — comes up everywhere: simultaneous equations, parametric problems, geometry coordinates.

Must learn

1.If two expressions both equal the same thing, set them equal to each other.

2.That's substitution.

YOU TRY · 5

If $x = 3p$ and $x = 7q$, write $p$ in terms of $q$.

Both equal $x$, so set them equal.

$3p = 7q$

$p = \dfrac{7}{3}q$

End of lesson

Algebra Manipulations — HL · Mathslive.ie