Simultaneous Equations — Mathslive.ie

Section 1 of 9

What are Simultaneous Equations?

Two equations, two unknowns. A single equation like $2x + 3y = 12$ has infinitely many $(x, y)$ pairs that work. Add a second equation and usually only one pair works for both.

Geometrically: each linear equation is a line. The solution is the point where the lines intersect.

Three types you must master

Type 1: $2 \times 2$ — two equations in $x$ and $y$
Type 2: $3 \times 3$ — three equations in $x$, $y$, $z$
Type 3: Substitution — one linear + one quadratic

The plan is always the same: get rid of one unknown so you have an equation with only one variable left. Solve that, then back-substitute.

You try

Does $(x, y) = (2, 3)$ satisfy both: $x + y = 5$ and $2x - y = 1$?

Sub $x=2$, $y=3$ into both equations. Do you get true statements?

First: $2 + 3 = 5$ ✓

Second: $2(2) - 3 = 4 - 3 = 1$ ✓

Yes — $(2, 3)$ is the solution.

Yes.

Section 2 of 9

Type 1: Basic $2 \times 2$

The plan: line up like terms in columns, then add or subtract the two equations to kill one variable.

Must learn

If the coefficients on one variable don't match, multiply one (or both) equations to make them match. Then add or subtract.

Worked example. Solve: $x + 2y = 3 \;\;$ and $\;\; 3x + y = 4$.

$\begin{aligned} x + 2y &= 3 \\ 3x + y &= 4 \end{aligned}$

Multiply the first by $-3$ to kill the $x$ column: $\;\;-3x - 6y = -9$

$\begin{aligned} -3x - 6y &= -9 \\ \phantom{-}3x + \phantom{1}y &= \phantom{-}4 \end{aligned}$ add the two

$-5y = -5 \;\;\Rightarrow\;\; y = 1$

Sub $y = 1$ into $x + 2y = 3$: $x + 2 = 3 \;\;\Rightarrow\;\; x = 1$

$x = 1, \;\; y = 1 \;\;$ — the lines meet at $(1, 1)$.

You try

Solve: $2x + y = 7 \;\;$ and $\;\; x - y = 2$

The $y$ coefficients are already $+1$ and $-1$ — perfect for adding.

$\begin{aligned} 2x + y &= 7 \\ \phantom{1}x - y &= 2 \end{aligned}$ (add)

$3x = 9 \;\;\Rightarrow\;\; x = 3$

Sub: $3 - y = 2 \;\;\Rightarrow\;\; y = 1$

$x = 3, \;\; y = 1$

You try

Solve: $3x + 2y = 16 \;\;$ and $\;\; 5x - 3y = 14$

Multiply first by $3$ and second by $2$ to make $y$ coefficients $\pm 6$.

$(1) \times 3:\;\; 9x + 6y = 48$

$(2) \times 2:\;\; 10x - 6y = 28$

Add: $\;19x = 76 \;\Rightarrow\; x = 4$

Sub into $(1)$: $12 + 2y = 16 \;\Rightarrow\; y = 2$

$x = 4, \;\; y = 2$

Section 3 of 9

$2 \times 2$ with Fractions

When fractions appear, your first job is to clear them — multiply each equation across by the lowest common denominator. Then it's a standard $2 \times 2$.

Worked example. Solve:

$\dfrac{x}{3} - \dfrac{y}{2} = 5 \;\;$ and $\;\; 2x - 1 = 5y$

Clear the first by $\times 6$ (LCM of 3 and 2):

$\dfrac{6x}{3} - \dfrac{6y}{2} = 30 \;\;\Rightarrow\;\; 2x - 3y = 30$

Tidy the second so $x$ and $y$ are on the left:

$2x - 1 = 5y \;\;\Rightarrow\;\; 2x - 5y = 1$

Now line them up and eliminate $x$:

$\begin{aligned} -2x + 3y &= -30 \\ \phantom{-}2x - 5y &= \phantom{-}1 \end{aligned}$ add

$-2y = -29 \;\;\Rightarrow\;\; y = \dfrac{29}{2}$

Sub into $2x - 3y = 30$:

$2x - 3\!\left(\dfrac{29}{2}\right) = 30$

Multiply across by $2$: $4x - 87 = 60$

$4x = 147 \;\;\Rightarrow\;\; x = \dfrac{147}{4}$

$x = \dfrac{147}{4}, \;\; y = \dfrac{29}{2}$

You try

Solve: $\dfrac{x}{2} + \dfrac{y}{3} = 4 \;\;$ and $\;\; x - y = 1$

Multiply the first equation by $6$ to clear the fractions.

First $\times 6$: $\;3x + 2y = 24$

Second $\times 2$: $\;2x - 2y = 2$

Add: $\;5x = 26 \;\Rightarrow\; x = \dfrac{26}{5}$

Sub: $\dfrac{26}{5} - y = 1 \;\Rightarrow\; y = \dfrac{26}{5} - 1 = \dfrac{21}{5}$

$x = \dfrac{26}{5}, \;\; y = \dfrac{21}{5}$

You try

Solve: $\dfrac{x + 1}{2} + \dfrac{y}{3} = 2 \;\;$ and $\;\; x + y = 3$

Multiply the first by $6$. Be careful expanding $\dfrac{6(x+1)}{2} = 3(x+1)$.

First $\times 6$: $\;3(x+1) + 2y = 12$

$\;3x + 3 + 2y = 12 \;\Rightarrow\; 3x + 2y = 9$

From second: $x = 3 - y$

Sub: $3(3 - y) + 2y = 9$

$9 - 3y + 2y = 9 \;\Rightarrow\; -y = 0 \;\Rightarrow\; y = 0$

$x = 3 - 0 = 3$

$x = 3, \;\; y = 0$

Section 4 of 9

Type 2: $3 \times 3$ Equations

Three equations, three unknowns. The strategy is to kill one variable twice — that gives you two equations in the remaining two unknowns. From there it's a standard $2 \times 2$.

The plan

Step 1: Pair up (i) & (ii) — eliminate $z$. Call the result (iv).
Step 2: Pair up (ii) & (iii) — eliminate $z$ again. Call this (v).
Step 3: Solve (iv) & (v) for $x$ and $y$.
Step 4: Sub back to find $z$.

Worked example. Solve:

$\begin{aligned} 2x + \phantom{1}y + \phantom{1}z &= 4 \;\;\;\;\text{(i)} \\ 3x + 5y - \phantom{1}z &= 7 \;\;\;\;\text{(ii)} \\ 5x - 2y + 2z &= 5 \;\;\;\;\text{(iii)} \end{aligned}$

Step 1. Eliminate $z$ from (i) and (ii) — the $z$ terms are $+z$ and $-z$, so just add:

$\begin{aligned} 2x + \phantom{1}y + z &= 4 \\ 3x + 5y - z &= 7 \end{aligned}$ add

$5x + 6y = 11 \;\;\;\;\text{(iv)}$

Step 2. Eliminate $z$ from (ii) and (iii). The $z$ terms are $-z$ and $+2z$ — multiply (ii) by $2$ first:

$\begin{aligned} \text{(ii)} \times 2:\;\; 6x + 10y - 2z &= 14 \\ \text{(iii)} \times 1:\;\; 5x - \phantom{1}2y + 2z &= \phantom{1}5 \end{aligned}$ add

$11x + 8y = 19 \;\;\;\;\text{(v)}$

Step 3. Now solve (iv) and (v). To kill $y$, multiply (iv) by $-4$ and (v) by $3$:

$\begin{aligned} \text{(iv)} \times (-4):\;\; -20x - 24y &= -44 \\ \text{(v)} \times 3:\;\; \phantom{-}33x + 24y &= \phantom{-}57 \end{aligned}$ add

$13x = 13 \;\;\Rightarrow\;\; x = 1$

Sub $x = 1$ into (iv): $5 + 6y = 11 \;\Rightarrow\; y = 1$

Step 4. Sub $x = 1$, $y = 1$ into (i):

$2 + 1 + z = 4 \;\;\Rightarrow\;\; z = 1$

$x = 1, \;\; y = 1, \;\; z = 1 \;\;\;$ — written as $(1, 1, 1)$.

Always check: sub your answer into the equation you didn't use at the end. Here, (iii): $5(1) - 2(1) + 2(1) = 5 - 2 + 2 = 5$ ✓

You try

Solve: $\begin{aligned} 2x + 3y + 4z &= 9 \;\;\;\text{(i)} \\ 3x - 2y + \phantom{1}z &= 2 \;\;\;\text{(ii)} \\ 5x + \phantom{1}y - 2z &= 4 \;\;\;\text{(iii)} \end{aligned}$

Kill $z$ twice. Pair (ii) & (iii) first — multiply (ii) by $2$ to match the $z$ coefficient in (iii).

(ii) $\times 2$: $\;6x - 4y + 2z = 4$

(iii) $\times 1$: $\;5x + y - 2z = 4$

Add: $\;11x - 3y = 8 \;\;\;\text{(iv)}$

(i) $\times 1$: $\;2x + 3y + 4z = 9$

(iii) $\times 2$: $\;10x + 2y - 4z = 8$

Add: $\;12x + 5y = 17 \;\;\;\text{(v)}$

(iv) $\times 5$: $\;55x - 15y = 40$

(v) $\times 3$: $\;36x + 15y = 51$

Add: $\;91x = 91 \;\Rightarrow\; x = 1$

Sub into (v): $\;12 + 5y = 17 \;\Rightarrow\; y = 1$

Sub into (ii): $\;3 - 2 + z = 2 \;\Rightarrow\; z = 1$

$(1, 1, 1)$

$x = 1, \;\; y = 1, \;\; z = 1$

You try

Solve: $\begin{aligned} 3x + \phantom{1}y + \phantom{1}z &= 0 \;\;\;\text{(i)} \\ \phantom{1}x - \phantom{1}y + \phantom{1}z &= 2 \;\;\;\text{(ii)} \\ 2x - 3y - \phantom{1}z &= 9 \;\;\;\text{(iii)} \end{aligned}$

The $z$ coefficients here are friendly. Pair (i) & (iii) — they cancel directly. Then pair (ii) & (iii).

(i) + (iii): $\;5x - 2y = 9 \;\;\;\text{(iv)}$

(ii) + (iii): $\;3x - 4y = 11 \;\;\;\text{(v)}$

(iv) $\times 2$: $\;10x - 4y = 18$

(v) $\times 1$: $\;3x - 4y = 11$

Subtract: $\;7x = 7 \;\Rightarrow\; x = 1$

Sub into (iv): $\;5 - 2y = 9 \;\Rightarrow\; y = -2$

Sub into (i): $\;3 + (-2) + z = 0 \;\Rightarrow\; z = -1$

$(1, -2, -1)$

$x = 1, \;\; y = -2, \;\; z = -1$

Section 5 of 9

Type 3: Substitution (Linear + Quadratic)

When one equation is linear and the other is quadratic, you can't eliminate by adding — the powers don't match. Use substitution instead.

Geometrically: a line meeting a circle (or other curve) — usually two points of intersection.

The plan

Step 1: Take the linear equation and isolate $y$ (or $x$).
Step 2: Sub into the quadratic. You'll get a quadratic in one variable.
Step 3: Solve. You usually get two values.
Step 4: Back-sub each value into the linear equation to pair them up.

Worked example 1. Solve: $x + y = 3 \;\;$ and $\;\; x^{2} + y^{2} = 5$

Step 1. From the linear: $y = 3 - x$

Step 2. Sub into the quadratic:

$x^{2} + (3 - x)^{2} = 5$

$x^{2} + 9 - 6x + x^{2} = 5$

$2x^{2} - 6x + 4 = 0$

$x^{2} - 3x + 2 = 0 \;\;$ (divide across by 2)

$(x - 2)(x - 1) = 0$

$x = 2 \;\;$ or $\;\; x = 1$

Step 4. Back-sub into $y = 3 - x$:

$x = 2 \;\Rightarrow\; y = 1 \;\;\;\;$ giving $\;(2, \, 1)$

$x = 1 \;\Rightarrow\; y = 2 \;\;\;\;$ giving $\;(1, \, 2)$

Two solutions: $(2, 1)$ and $(1, 2)$.

Why two solutions? The line $x + y = 3$ cuts the circle $x^{2} + y^{2} = 5$ at two points.

Worked example 2. Solve: $2x + 3y = 5 \;\;$ and $\;\; x^{2} + y^{2} = 2$

Isolate $y$ from the linear:

$3y = 5 - 2x \;\;\Rightarrow\;\; y = \dfrac{5 - 2x}{3}$

Sub into the quadratic:

$x^{2} + \!\left(\dfrac{5 - 2x}{3}\right)^{\!2} = 2$

$x^{2} + \dfrac{(5 - 2x)^{2}}{9} = 2$

$x^{2} + \dfrac{25 - 20x + 4x^{2}}{9} = 2$

Clear the fraction by $\times 9$:

$9x^{2} + 25 - 20x + 4x^{2} = 18$

$13x^{2} - 20x + 7 = 0$

Factor: looking for two numbers multiplying to $13 \times 7 = 91$ and adding to $-20$: that's $-13$ and $-7$.

$13x^{2} - 13x - 7x + 7 = 0$

$13x(x - 1) - 7(x - 1) = 0$

$(x - 1)(13x - 7) = 0$

$x = 1 \;\;$ or $\;\; x = \dfrac{7}{13}$

Back-sub into $y = \dfrac{5 - 2x}{3}$:

$x = 1 \;\Rightarrow\; y = \dfrac{5 - 2}{3} = 1$

$x = \dfrac{7}{13} \;\Rightarrow\; y = \dfrac{5 - \dfrac{14}{13}}{3} = \dfrac{\dfrac{51}{13}}{3} = \dfrac{17}{13}$

Two solutions: $(1, 1)$ and $\!\left(\dfrac{7}{13}, \dfrac{17}{13}\right)$

You try

Solve: $3x - y = 8 \;\;$ and $\;\; x^{2} + y^{2} = 10$

From the linear, isolate $y$: $\;y = 3x - 8$. Sub into the quadratic.

$y = 3x - 8$

$x^{2} + (3x - 8)^{2} = 10$

$x^{2} + 9x^{2} - 48x + 64 = 10$

$10x^{2} - 48x + 54 = 0$

$5x^{2} - 24x + 27 = 0 \;$ ($\div 2$)

$5x^{2} - 15x - 9x + 27 = 0$

$5x(x - 3) - 9(x - 3) = 0$

$(x - 3)(5x - 9) = 0$

$x = 3 \;\;$ or $\;\; x = \dfrac{9}{5}$

$x = 3 \;\Rightarrow\; y = 9 - 8 = 1$

$x = \dfrac{9}{5} \;\Rightarrow\; y = 3\!\left(\dfrac{9}{5}\right) - 8 = \dfrac{27}{5} - \dfrac{40}{5} = -\dfrac{13}{5}$

$(3, 1)$ and $\!\left(\dfrac{9}{5}, \, -\dfrac{13}{5}\right)$

You try

Solve: $x - y = 1 \;\;$ and $\;\; x^{2} + y^{2} = 13$

$y = x - 1$, then sub.

$y = x - 1$

$x^{2} + (x - 1)^{2} = 13$

$x^{2} + x^{2} - 2x + 1 = 13$

$2x^{2} - 2x - 12 = 0$

$x^{2} - x - 6 = 0$

$(x - 3)(x + 2) = 0$

$x = 3 \;\Rightarrow\; y = 2$

$x = -2 \;\Rightarrow\; y = -3$

$(3, 2)$ and $(-2, -3)$

You try

Solve: $x + 2y = 5 \;\;$ and $\;\; x^{2} + y^{2} = 10$

Isolate $x$ this time: $x = 5 - 2y$. Either variable works — pick the one that's easier.

$x = 5 - 2y$

$(5 - 2y)^{2} + y^{2} = 10$

$25 - 20y + 4y^{2} + y^{2} = 10$

$5y^{2} - 20y + 15 = 0$

$y^{2} - 4y + 3 = 0$

$(y - 1)(y - 3) = 0$

$y = 1 \;\Rightarrow\; x = 3$

$y = 3 \;\Rightarrow\; x = -1$

$(3, 1)$ and $(-1, 3)$

Section 6 of 9

Sim Equations with Exponents

Sometimes the unknowns appear inside powers — this comes up in geometric series problems where you're given $ar^{2}$ and $ar^{4}$, for example.

The trick: isolate one variable from the simpler equation, then sub into the other. Powers will then collapse using the index rule $\dfrac{r^{m}}{r^{n}} = r^{m - n}$.

Worked example. Solve: $ar^{2} = 12 \;\;$ and $\;\; ar^{4} = 48$

Isolate $a$ from the first:

$a = \dfrac{12}{r^{2}}$

Sub into the second:

$\dfrac{12}{r^{2}} \cdot r^{4} = 48$

$12 r^{2} = 48 \;\;$ (using $\dfrac{r^{4}}{r^{2}} = r^{2}$)

$r^{2} = 4 \;\;\Rightarrow\;\; r = \pm 2$

Back to $a$:

$a = \dfrac{12}{r^{2}} = \dfrac{12}{4} = 3$

$a = 3, \;\; r = \pm 2$

Shortcut you'll spot: dividing the second by the first instantly gives $\dfrac{ar^{4}}{ar^{2}} = \dfrac{48}{12} \;\Rightarrow\; r^{2} = 4$. Use whichever method you prefer.

You try

Solve: $ar = 6 \;\;$ and $\;\; ar^{3} = 54$

Isolate $a = \dfrac{6}{r}$ and sub, or divide the second by the first.

$a = \dfrac{6}{r}$

$\dfrac{6}{r} \cdot r^{3} = 54$

$6 r^{2} = 54 \;\Rightarrow\; r^{2} = 9 \;\Rightarrow\; r = \pm 3$

$a = \dfrac{6}{r}$: $\;r = 3 \;\Rightarrow\; a = 2$; $\;r = -3 \;\Rightarrow\; a = -2$

$(a, r) = (2, 3) \;$ or $\;(-2, -3)$

Section 7 of 9

"Become Same" — Comparing Coefficients

An identity is an equation that holds for every value of $x$. When you're told an expression is true "for all $x$", a quick comparison of coefficients turns the problem into simultaneous equations.

The plan

Step 1: Multiply out the left-hand side fully.
Step 2: Match the $x$ coefficients on each side.
Step 3: Match the constant terms on each side.
Step 4: You now have a $2 \times 2$ — solve it.

Worked example. Given $p(x + 1) + q(3x - 5) = 5x + 6 \;\;$ for all $x$, find $p$ and $q$.

Step 1. Multiply out:

$px + p + 3qx - 5q = 5x + 6$

$(p + 3q)x + (p - 5q) = 5x + 6$

Steps 2 & 3. Match coefficients:

$x$ coefficient: $\;\;p + 3q = 5 \;\;\;\;\text{(i)}$

constant: $\;\;p - 5q = 6 \;\;\;\;\text{(ii)}$

Step 4. Subtract (ii) from (i):

$\begin{aligned} \phantom{-}p + 3q &= \phantom{-}5 \\ -(p - 5q) &= -6 \end{aligned}$

$8q = -1 \;\;\Rightarrow\;\; q = -\dfrac{1}{8}$

Sub into (i):

$p + 3\!\left(-\dfrac{1}{8}\right) = 5$

$p - \dfrac{3}{8} = 5$

$p = 5 + \dfrac{3}{8} = \dfrac{40}{8} + \dfrac{3}{8} = \dfrac{43}{8}$

$p = \dfrac{43}{8}, \;\; q = -\dfrac{1}{8}$

You try

Given $a(x + 3) + b(2x + 1) = 5x + 7 \;\;$ for all $x$, find $a$ and $b$.

Multiply out the left side, group $x$ terms and constants, then match.

$ax + 3a + 2bx + b = 5x + 7$

$(a + 2b)x + (3a + b) = 5x + 7$

$a + 2b = 5 \;\;\;\text{(i)}$

$3a + b = 7 \;\;\;\text{(ii)}$

(ii) $\times 2$: $\;6a + 2b = 14$

Subtract (i): $\;5a = 9 \;\Rightarrow\; a = \dfrac{9}{5}$

From (i): $\;b = \dfrac{5 - a}{2} \cdot \dfrac{1}{1}$... easier: $b = \dfrac{5 - \dfrac{9}{5}}{2} = \dfrac{\dfrac{16}{5}}{2} = \dfrac{8}{5}$

$a = \dfrac{9}{5}, \;\; b = \dfrac{8}{5}$

You try

Given $a(x - 2) + b(x + 4) = 4x + 2 \;\;$ for all $x$, find $a$ and $b$.

Multiply out and match coefficients.

$ax - 2a + bx + 4b = 4x + 2$

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

$a + b = 4 \;\;\;\text{(i)}$

$-2a + 4b = 2 \;\;\;\text{(ii)}$

(i) $\times 2$: $\;2a + 2b = 8$

Add to (ii): $\;6b = 10 \;\Rightarrow\; b = \dfrac{5}{3}$

$a = 4 - \dfrac{5}{3} = \dfrac{12}{3} - \dfrac{5}{3} = \dfrac{7}{3}$

$a = \dfrac{7}{3}, \;\; b = \dfrac{5}{3}$

Section 8 of 9

Worded Problems

The maths is the same — but you have to set up the equations yourself from a story.

The plan

Step 1: Pick what each unknown represents — $x = \ldots$, $y = \ldots$
Step 2: Write one equation per fact you're given.
Step 3: Solve as a normal $2 \times 2$.
Step 4: Re-read the question — answer in words, not just numbers.

Worked example 1 — the farm. A farm has pigs and hens. There are $32$ heads in total and $104$ legs. How many of each?

Step 1. Let $\;x = $ number of pigs, $\;y = $ number of hens.

Step 2. Each animal has one head:

Heads: $\;\;x + y = 32$

Pigs have $4$ legs, hens have $2$:

Legs: $\;\;4x + 2y = 104$

Step 3. Multiply the first by $-2$:

$\begin{aligned} -2x - 2y &= -64 \\ \phantom{-}4x + 2y &= \phantom{-}104 \end{aligned}$ add

$2x = 40 \;\;\Rightarrow\;\; x = 20$

$y = 32 - 20 = 12$

$20$ pigs and $12$ hens.

Worked example 2 — the wallet. I have €$170$ made up of €$10$ and €$20$ notes. I have $11$ notes in total. How many of each?

Let $\;x = $ number of €$10$ notes, $\;y = $ number of €$20$ notes.

Notes: $\;\;x + y = 11$

Value: $\;\;10x + 20y = 170$

Multiply the first by $-10$:

$\begin{aligned} -10x - 10y &= -110 \\ \phantom{-}10x + 20y &= \phantom{-}170 \end{aligned}$ add

$10y = 60 \;\;\Rightarrow\;\; y = 6$

$x = 11 - 6 = 5$

$5$ notes of €$10$ and $6$ notes of €$20$.

You try

I have $25$ coins, either €$1$ or €$2$, with a total value of €$42$. How many of each?

Let $x = $ number of €$1$ coins, $y = $ number of €$2$ coins. One equation for the count, one for the value.

Coins: $\;x + y = 25$

Value: $\;x + 2y = 42$

Subtract first from second: $\;y = 17$

$x = 25 - 17 = 8$

$8$ coins of €$1$ and $17$ coins of €$2$.

$8$ coins of €$1$, $17$ coins of €$2$.

You try

A cinema sells adult tickets at €$12$ and child tickets at €$7$. One night they sold $50$ tickets for a total of €$475$. How many of each were sold?

Let $x = $ adult tickets, $y = $ child tickets. Tickets equation: $x + y = 50$. Money equation: $12x + 7y = 475$.

$x + y = 50 \;\;\;\text{(i)}$

$12x + 7y = 475 \;\;\;\text{(ii)}$

(i) $\times -7$: $\;-7x - 7y = -350$

Add to (ii): $\;5x = 125 \;\Rightarrow\; x = 25$

$y = 50 - 25 = 25$

$25$ adult tickets and $25$ child tickets.

$25$ adult and $25$ child.

You try

Three coffees and two scones cost €$13.50$. Two coffees and three scones cost €$12$. Find the price of one coffee and one scone.

Let $c = $ price of a coffee, $s = $ price of a scone. Write the two equations, then eliminate.

$3c + 2s = 13.50 \;\;\;\text{(i)}$

$2c + 3s = 12 \;\;\;\text{(ii)}$

(i) $\times 3$: $\;9c + 6s = 40.50$

(ii) $\times 2$: $\;4c + 6s = 24$

Subtract: $\;5c = 16.50 \;\Rightarrow\; c = 3.30$

From (i): $\;2s = 13.50 - 3(3.30) = 13.50 - 9.90 = 3.60$

$s = 1.80$

Coffee €$3.30$, scone €$1.80$.

Section 9 of 9

Recap — One Strategy Per Type

$2 \times 2$ — elimination

Make a coefficient match (multiply if needed). Add or subtract to kill one variable. Solve for the other. Back-sub.

$2 \times 2$ with fractions

Clear the fractions first by multiplying across by the LCM. Then it's a normal $2 \times 2$.

$3 \times 3$ — double elimination

Kill the same variable from two different pairs. That gives you a $2 \times 2$ in the other two. Solve, then back-sub for the third.

Linear + quadratic — substitution

Isolate $y$ (or $x$) from the linear equation. Sub into the quadratic. Solve. Expect two solutions — back-sub each.

Exponent-style

Isolate one variable from the simpler equation. Sub in. The index rule $\dfrac{r^{m}}{r^{n}} = r^{m - n}$ does the heavy lifting.

"Become Same"

Multiply out. Match $x$ coefficients. Match constants. That's your $2 \times 2$.

Worded

Name your unknowns. One equation per fact. Solve. Answer in words.

That's Simultaneous Equations.

You can now solve $2 \times 2$, $3 \times 3$, linear-plus-quadratic, exponent pairs, identity matching, and word problems — all using the same core idea: kill one unknown, then back-substitute.