Solving simultaneous equations

Two unknowns, two clues — and only one pair of numbers satisfies both at once. That is the whole idea behind simultaneous linear equations: two equations in $x$ and $y$ , each drawing a straight line, and the answer is the single point where both are true together. ICSE Class 9 asks you to find that point without drawing anything, using pure algebra. This is the second big strand of the Algebra chapter, right after factorisation, and it is scoped tightly: numerical coefficients only, exactly two variables, and three named algebraic routes to the answer — substitution, elimination, and cross-multiplication. (Plotting the lines belongs to Coordinate Geometry, not here.) Master these three routes and you can solve any well-posed pair the exam throws at you, check your own work by a second method, and set up equations from a word problem involving two related unknown quantities.

The core insight

A pair of linear equations in $x$ and $y$ has exactly one common solution $(x,y)$ , found algebraically by substitution, elimination, or cross-multiplication.

The core idea

A pair of simultaneous linear equations in two variables has the general form

a_1x+b_1y=c_1, \qquad a_2x+b_2y=c_2

where: $a_1,b_1,c_1,a_2,b_2,c_2$ are given numbers (numerical coefficients — no letters), and $x,y$ are the two unknowns we must find together, as one matching pair.

Substitution isolates one variable from one equation, then plugs that expression into the other equation, collapsing two unknowns into one.

Elimination scales one or both equations so that the coefficient of one variable matches (or is equal-and-opposite) in both, then adds or subtracts the equations to remove that variable entirely.

Cross-multiplication skips the scaling step altogether. Write both equations in standard form $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ , and read the solution straight off a fixed formula:

\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}

where: each numerator is a cross-product of coefficients taken in a cyclic pattern. A common memory aid is to write $b_1,c_1,a_1,b_1,c_1$ above and $b_2,c_2,a_2,b_2,c_2$ below, in a row, and cross-multiply neighbouring pairs — this rebuilds the three expressions above without memorising subscripts. Once $x$ and $y$ are each expressed as a ratio to $1$ , both unknowns come out in one pass, with no intermediate scaling of either equation — a real advantage when the coefficients are awkward numbers that would otherwise force ugly fractions under elimination.

Figure

Three paths from two equations converging on one solution point

Choose the method by what the equations look like. Substitution is fastest when a variable already has coefficient $1$ or is trivial to isolate. Elimination is fastest when coefficients are close to matching, or become so after multiplying by small whole numbers. Cross-multiplication is the mechanical fallback that always works — provided $a_1b_2-a_2b_1 \neq 0$ ; if that quantity is zero, the lines are parallel or coincident and there is no single unique solution, a case beyond what Class 9 numerical problems normally test. Word problems are simply simultaneous equations wearing a disguise. Two related unknown quantities — an age and a friend's age, the cost of two kinds of fruit, the speed of a boat with and against a current — turn into two equations the moment the problem states two separate numerical facts connecting them. The skill is translating each sentence of the problem into one equation, keeping the same two letters for the same two unknowns throughout, and only then choosing which of the three methods to solve with. Whatever method is used, the final ordered pair should always be checked against both original equations, not just re-derived once and trusted.

Derivation AWhere the cross-multiplication formula comes from

The goal is to show that the cross-multiplication rule is not a magic recipe but ordinary elimination carried out once, in general, on the standard-form pair $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ — so that the formula can be trusted and reproduced rather than memorised blindly, and so that the appearance of $a_1b_2-a_2b_1$ in the denominator is understood as the same "coefficients must not already match" condition that elimination itself needs.

Step 1. Eliminate $y$ first. Multiply the first equation by $b_2$ and the second by $b_1$ , so the $y$ -coefficient becomes $b_1b_2$ in both: $a_1b_2x+b_1b_2y+b_2c_1=0$ and $a_2b_1x+b_1b_2y+b_1c_2=0$ .

Step 2. Subtract the second new equation from the first; the $b_1b_2y$ terms cancel exactly: $(a_1b_2-a_2b_1)x+(b_2c_1-b_1c_2)=0$ , which rearranges to $(a_1b_2-a_2b_1)x=b_1c_2-b_2c_1$ , so $x=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}$ — the first ratio in the formula, provided the denominator is not zero.

Step 3. Eliminate $x$ by the same idea, multiplying by $a_2$ and $a_1$ this time and subtracting: this produces $(a_1b_2-a_2b_1)y=c_1a_2-c_2a_1$ , so $y=\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}$ — the second ratio, sharing the same denominator as $x$ .

Step 4. Write both results as one chain. Since $x$ and $y$ share the same denominator $a_1b_2-a_2b_1$ , and that denominator equals the value that $1$ is set equal to, the two results combine into the compact three-way ratio $\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}$ — exactly the cross-multiplication formula, now seen as elimination performed in general once and for all.

Worked example

Solve the pair $2x+3y=13$ and $3x-2y=0$ by elimination, then confirm with cross-multiplication.

Step 1. Match a coefficient. Multiply the first equation by $2$ and the second by $3$ , so the $y$ -coefficients become equal in size: $4x+6y=26$ and $9x-6y=0$ .

Step 2. Add the two scaled equations, since the $y$ -terms are opposite in sign: $13x=26$ , so $x=2$ .

Step 3. Substitute $x=2$ into the other original equation, $3x-2y=0$ : $3(2)-2y=0 \implies 6=2y \implies y=3$ .

Step 4. Check in the first original equation: $2(2)+3(3)=4+9=13$ . Correct, so $x=2,y=3$ . Notice the check used the equation that was not used to find $y$ in Step 3 — that is what makes it a genuine check rather than a repeat of the same working.

Competitive stretch

Confirm the same pair by cross-multiplication. Rewrite in standard form: $2x+3y-13=0$ and $3x-2y-0=0$ , so $a_1=2,b_1=3,c_1=-13$ and $a_2=3,b_2=-2,c_2=0$ . Then $\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{x}{(3)(0)-(-2)(-13)}=\dfrac{x}{-26}$ , $\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{y}{(-13)(3)-(0)(2)}=\dfrac{y}{-39}$ , and $\dfrac{1}{a_1b_2-a_2b_1}=\dfrac{1}{(2)(-2)-(3)(3)}=\dfrac{1}{-13}$ . So $x=-26/-13=2$ and $y=-39/-13=3$ . The solution is confirmed: $x=2,\ y=3$ .

Common traps

Forgetting to multiply every term of an equation, including the constant on the right, when scaling it for elimination — this quietly breaks the equation and gives a wrong final pair.
Sign errors when adding or subtracting during elimination — always check, from the signs of the matched coefficients, whether addition or subtraction removes the variable.
Substituting the found value back into the same equation used to derive it rather than the other equation, which can hide an arithmetic error instead of catching it.
Misplacing coefficients in the cross-multiplication cyclic pattern, or forgetting to rearrange an equation like $2x+3y=13$ into standard form $2x+3y-13=0$ first, so that $c_1=-13$ and not $13$ . In a word problem, the matching error is letting the two unknowns drift — reusing $x$ for a different quantity partway through — which produces two equations that no longer describe the same pair of unknowns.

Ask the tutor: Not sure which method is fastest for a tricky pair of equations — ask your AI tutor to solve the same problem by all three methods side by side and compare the steps.

Why do I get a different-looking answer if I use elimination versus cross-multiplication on the same problem?

You should not — a correctly solved pair gives the same $(x,y)$ regardless of method, because all three routes are just different ways of combining the same two equations; the worked example above shows elimination and cross-multiplication landing on identical numbers for the same pair. If your answers disagree, an arithmetic slip happened somewhere, most often a sign error while adding or subtracting scaled equations, or a coefficient misplaced in the cross-multiplication pattern. Solving by two methods is actually the recommended way to self-check tricky problems, especially in the exam's final minutes when time allows a quick second pass.

Do I always have to rearrange equations into standard form before cross-multiplying?

Yes. The formula $\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}$ is defined for equations already written as $a_1x+b_1y+c_1=0$ , with everything moved to one side and set equal to zero. An equation like $2x+3y=13$ must first become $2x+3y-13=0$ , giving $c_1=-13$ , not $13$ . Skipping this step is the single most common reason students plug numbers into the formula and get a wrong sign, since the constant term silently flips sign the moment it crosses the equals sign.

How do I know which of the three methods to use in an exam?

Look at the equations for ten seconds before writing anything. If a variable already stands alone or has coefficient $1$ , substitution is fastest, since there is nothing to scale before substituting. If the coefficients are equal, or become equal after multiplying by small whole numbers like $2$ or $3$ , elimination is quickest and avoids fractions creeping into the working. If neither shortcut is obvious, cross-multiplication is the reliable fallback — it always works as long as $a_1b_2-a_2b_1 \neq 0$ , though it takes a little longer to set up because both equations must first be rearranged into standard form.

Solving simultaneous equations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads