Pair of Linear Equations in Two Variables • Topic 2 of 3

Algebraic Methods: Substitution, Elimination, Cross-Multiplication

Why algebraic methods? Graphs are intuitive but only approximate when the solution is not a "nice" integer point. Algebraic methods give exact answers. The three methods in the syllabus are substitution, elimination and cross-multiplication.

1. Substitution method. Steps: (i) from one equation express one variable in terms of the other; (ii) substitute that expression into the second equation to get a single-variable equation; (iii) solve it; (iv) back-substitute to find the second variable. Best when one coefficient is already \(1\) or \(-1\).

Example: solve \(2x+y=7\) and \(x-y=2\). From the second, \(x=y+2\). Substitute: \(2(y+2)+y=7\Rightarrow 3y+4=7\Rightarrow y=1\), then \(x=3\).

2. Elimination method. Steps: (i) multiply one or both equations by suitable numbers so that the coefficients of one chosen variable become numerically equal; (ii) add or subtract the equations to eliminate that variable; (iii) solve for the remaining variable; (iv) substitute back. Choose to add when the equal coefficients have opposite signs, and subtract when they have the same sign. Best when neither coefficient is \(1\).

3. Cross-multiplication method. Write both equations in standard form \(a_{1}x + b_{1}y + c_{1} = 0\) and \(a_{2}x + b_{2}y + c_{2} = 0\). Then

\(\dfrac{x}{b_{1}c_{2}-b_{2}c_{1}}=\dfrac{y}{c_{1}a_{2}-c_{2}a_{1}}=\dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}\).

From this, \(x=\dfrac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) and \(y=\dfrac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}\), provided \(a_{1}b_{2}-a_{2}b_{1}\ne 0\).

A memory device for cross-multiplication. Write the coefficients in two rows, starting and ending with the \(b\)-column, and cross-multiply along the diagonals:

\(\begin{array}{cccc} b_{1} & c_{1} & a_{1} & b_{1}\\ b_{2} & c_{2} & a_{2} & b_{2}\end{array}\)

Number-of-solutions test using the cross-multiplication denominator. If \(a_{1}b_{2}-a_{2}b_{1}\ne 0\) the system has a unique solution. If \(a_{1}b_{2}-a_{2}b_{1}=0\) the lines are parallel or coincident: compare \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\) with \(\dfrac{c_{1}}{c_{2}}\) — equal gives infinitely many, unequal gives none.

Equations reducible to linear form. Some equations are not linear in \(x,y\) but become linear after a substitution. For instance with \(\dfrac{2}{x}+\dfrac{3}{y}=13\) and \(\dfrac{5}{x}-\dfrac{4}{y}=-2\), put \(u=\dfrac1x,\ v=\dfrac1y\); the system becomes \(2u+3v=13,\ 5u-4v=-2\), which is linear in \(u,v\). Solve, then recover \(x=\dfrac1u,\ y=\dfrac1v\). The same trick handles upstream/downstream boat problems.

Common mistakes. (1) In elimination, subtracting without changing every sign of the second row. (2) In cross-multiplication, mixing up the order \(b_{1}c_{2}-b_{2}c_{1}\) (keep the subscripts patterned). (3) Forgetting to convert to standard \(=0\) form, so a sign on \(c\) is wrong. (4) After a substitution like \(u=1/x\), forgetting to convert \(u\) back to \(x\). (5) Dividing by \(a_{1}b_{2}-a_{2}b_{1}\) when it is zero — that signals no unique solution, not an arithmetic slip.

Which method should I pick? If one variable already has coefficient \(1\) or \(-1\), substitution is quickest. If both equations have awkward coefficients of the same variable, elimination usually wins. Cross-multiplication is a one-shot formula method — useful in exams once the equations are in standard form, and it doubles as a quick way to read off the number of solutions from the denominator. All three give exactly the same answer; only the labour differs.

Always verify. Substitute the values you obtain back into both original equations. A single check catches sign slips and arithmetic errors, which are the most common source of lost marks in this chapter.

Side-by-side flow of the substitution and elimination methods Two equations in x and y — two routes SUBSTITUTION 1. Solve one eqn for x (or y) 2. Substitute into the other 3. Solve the single variable 4. Back-substitute for the rest ELIMINATION 1. Make one coefficient equal 2. Add / subtract to drop a variable 3. Solve the single variable 4. Substitute back for the rest Both routes reach the same (x, y) e.g. 2x + y = 7, x − y = 2 → x = 3, y = 1 Cross-multiplication formula laid out as a schematic Cross-multiplication shortcut x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁) For 2x + y − 7 = 0 and x − y − 2 = 0: x/(−9) = y/(−3) = 1/(−3) → x = 3, y = 1
Solved Examples
1
Worked Example
Solve by substitution: \(3x + 2y = 12\) and \(x - y = -1\).
Solution
  1. Step 1: From \(x-y=-1\), express \(x=y-1\).
  2. Step 2: Substitute into \(3x+2y=12\): \(3(y-1)+2y=12\).
  3. Step 3: \(3y-3+2y=12\Rightarrow 5y=15\Rightarrow y=3\).
  4. Step 4: Back-substitute: \(x=3-1=2\).
  5. Step 5: Check: \(3(2)+2(3)=12\) and \(2-3=-1\).

Answer: \(x=2,\ y=3\).

2
Worked Example
Solve by elimination: \(4x + 3y = 18\) and \(5x - 2y = 11\).
Solution
  1. Step 1: Make \(y\)-coefficients equal: multiply first by \(2\), second by \(3\): \(8x+6y=36\) and \(15x-6y=33\).
  2. Step 2: Add (opposite signs on \(y\)): \(23x=69\Rightarrow x=3\).
  3. Step 3: Substitute \(x=3\) into \(4x+3y=18\): \(12+3y=18\Rightarrow 3y=6\Rightarrow y=2\).
  4. Step 4: Check: \(5(3)-2(2)=15-4=11\).

Answer: \(x=3,\ y=2\).

3
Worked Example
Solve by cross-multiplication: \(2x + 5y = 1\) and \(3x + 2y = 7\).
Solution
  1. Step 1: Standard form: \(2x+5y-1=0\) and \(3x+2y-7=0\); so \(a_{1}=2,b_{1}=5,c_{1}=-1,\ a_{2}=3,b_{2}=2,c_{2}=-7\).
  2. Step 2: Denominator \(a_{1}b_{2}-a_{2}b_{1}=2(2)-3(5)=4-15=-11\).
  3. Step 3: \(x\)-numerator \(b_{1}c_{2}-b_{2}c_{1}=5(-7)-2(-1)=-35+2=-33\).
  4. Step 4: \(y\)-numerator \(c_{1}a_{2}-c_{2}a_{1}=(-1)(3)-(-7)(2)=-3+14=11\).
  5. Step 5: \(x=\dfrac{-33}{-11}=3,\quad y=\dfrac{11}{-11}=-1\).

Answer: \(x=3,\ y=-1\).

4
Worked Example
Solve by substitution: \(0.2x + 0.3y = 1.3\) and \(0.4x + 0.5y = 2.3\).
Solution
  1. Step 1: Multiply through by \(10\): \(2x+3y=13\) and \(4x+5y=23\).
  2. Step 2: From the first, \(x=\dfrac{13-3y}{2}\).
  3. Step 3: Substitute: \(4\cdot\dfrac{13-3y}{2}+5y=23\Rightarrow 2(13-3y)+5y=23\).
  4. Step 4: \(26-6y+5y=23\Rightarrow -y=-3\Rightarrow y=3\); then \(x=\dfrac{13-9}{2}=2\).

Answer: \(x=2,\ y=3\).

5
Worked Example
Solve by elimination: \(\dfrac{x}{2}+\dfrac{2y}{3}=-1\) and \(x-\dfrac{y}{3}=3\).
Solution
  1. Step 1: Clear fractions. First \(\times 6\): \(3x+4y=-6\). Second \(\times 3\): \(3x-y=9\).
  2. Step 2: Subtract (same \(3x\)): \((3x+4y)-(3x-y)=-6-9\Rightarrow 5y=-15\Rightarrow y=-3\).
  3. Step 3: Substitute into \(3x-y=9\): \(3x+3=9\Rightarrow 3x=6\Rightarrow x=2\).

Answer: \(x=2,\ y=-3\).

6
Worked Example
Solve \(x + y = 14\) and \(x - y = 4\) by both elimination and substitution.
Solution
  1. Step 1 (elimination): Add the equations: \(2x=18\Rightarrow x=9\).
  2. Step 2: Subtract: \(2y=10\Rightarrow y=5\).
  3. Step 3 (substitution check): from \(x-y=4\), \(x=y+4\); into \(x+y=14\): \(2y+4=14\Rightarrow y=5,\ x=9\).

Answer: \(x=9,\ y=5\).

7
Worked Example
Solve the pair reducible to linear form: \(\dfrac{2}{x}+\dfrac{3}{y}=13\) and \(\dfrac{5}{x}-\dfrac{4}{y}=-2\).
Solution
  1. Step 1: Let \(u=\dfrac1x,\ v=\dfrac1y\). Equations: \(2u+3v=13\) and \(5u-4v=-2\).
  2. Step 2: Multiply first by \(4\), second by \(3\): \(8u+12v=52\) and \(15u-12v=-6\).
  3. Step 3: Add: \(23u=46\Rightarrow u=2\); then \(2(2)+3v=13\Rightarrow 3v=9\Rightarrow v=3\).
  4. Step 4: Recover \(x=\dfrac1u=\dfrac12,\ y=\dfrac1v=\dfrac13\).

Answer: \(x=\tfrac12,\ y=\tfrac13\).

8
Worked Example
For what value of \(k\) does \(2x + 3y = 7\) and \(6x + (k+2)y = 22\) have no solution?
Solution
  1. Step 1: No solution needs \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\ne\dfrac{c_{1}}{c_{2}}\).
  2. Step 2: \(\dfrac{2}{6}=\dfrac13\), so set \(\dfrac{3}{k+2}=\dfrac13\Rightarrow k+2=9\Rightarrow k=7\).
  3. Step 3: Check constants: \(\dfrac{c_{1}}{c_{2}}=\dfrac{-7}{-22}=\dfrac{7}{22}\ne\dfrac13\), so they are parallel.

Answer: \(k=7\) (parallel lines, no solution).

9
Worked Example
Solve by cross-multiplication: \(8x + 5y = 9\) and \(3x + 2y = 4\).
Solution
  1. Step 1: Standard form: \(8x+5y-9=0,\ 3x+2y-4=0\). So \(a_{1}=8,b_{1}=5,c_{1}=-9,\ a_{2}=3,b_{2}=2,c_{2}=-4\).
  2. Step 2: Denominator \(a_{1}b_{2}-a_{2}b_{1}=8(2)-3(5)=16-15=1\).
  3. Step 3: \(x\)-num \(=b_{1}c_{2}-b_{2}c_{1}=5(-4)-2(-9)=-20+18=-2\).
  4. Step 4: \(y\)-num \(=c_{1}a_{2}-c_{2}a_{1}=(-9)(3)-(-4)(8)=-27+32=5\).
  5. Step 5: \(x=\dfrac{-2}{1}=-2,\ y=\dfrac{5}{1}=5\).

Answer: \(x=-2,\ y=5\).

10
Worked Example
Solve by elimination: \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\).
Solution
  1. Step 1: Rewrite: \(3x-5y=4\) and \(9x-2y=7\).
  2. Step 2: Multiply first by \(3\): \(9x-15y=12\).
  3. Step 3: Subtract the second: \((9x-15y)-(9x-2y)=12-7\Rightarrow -13y=5\Rightarrow y=-\dfrac{5}{13}\).
  4. Step 4: Substitute into \(3x-5y=4\): \(3x+\dfrac{25}{13}=4\Rightarrow 3x=\dfrac{27}{13}\Rightarrow x=\dfrac{9}{13}\).

Answer: \(x=\tfrac{9}{13},\ y=-\tfrac{5}{13}\).

11
Worked Example
Use cross-multiplication to decide the nature of \(2x + 3y = 5\) and \(4x + 6y = 9\).
Solution
  1. Step 1: \(a_{1}=2,b_{1}=3,a_{2}=4,b_{2}=6\); denominator \(a_{1}b_{2}-a_{2}b_{1}=2(6)-4(3)=12-12=0\).
  2. Step 2: Denominator zero ⇒ not a unique solution; check constants.
  3. Step 3: \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac12\) but \(\dfrac{c_{1}}{c_{2}}=\dfrac{-5}{-9}=\dfrac59\ne\dfrac12\).
  4. Step 4: Coefficient ratios equal, constant ratio different ⇒ parallel.

Answer: No solution (parallel lines).

12
Worked Example
Solve the pair reducible to linear form: \(\dfrac{1}{2x}+\dfrac{1}{3y}=2\) and \(\dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}\).
Solution
  1. Step 1: Let \(u=\dfrac1x,\ v=\dfrac1y\): \(\dfrac{u}{2}+\dfrac{v}{3}=2\) and \(\dfrac{u}{3}+\dfrac{v}{2}=\dfrac{13}{6}\).
  2. Step 2: Multiply both by \(6\): \(3u+2v=12\) and \(2u+3v=13\).
  3. Step 3: \((3u+2v)\times3-(2u+3v)\times2:\ 9u+6v-4u-6v=36-26\Rightarrow 5u=10\Rightarrow u=2\).
  4. Step 4: Then \(3(2)+2v=12\Rightarrow v=3\). Recover \(x=\dfrac12,\ y=\dfrac13\).

Answer: \(x=\tfrac12,\ y=\tfrac13\).

Key Points

  • Substitution: Isolate one variable, substitute into the other equation
  • Elimination: Make coefficients equal, add/subtract to eliminate a variable
  • Cross-multiplication: Direct formula when equations are in standard form
  • All three methods give the same answer — choose the easiest for each problem
  • Cross-multiplication fails when \(a_{1}b_{2} - a_{2}b_{1}\) = 0 (parallel or coincident lines)
  • Always check your answer in both original equations
  • Add equations when equal coefficients have opposite signs; subtract when they have the same sign
  • Cross-multiplication denominator \(a_{1}b_{2}-a_{2}b_{1}\ne0 \Rightarrow\) unique solution
  • Equations in \(\tfrac1x,\tfrac1y\) become linear via \(u=\tfrac1x,\ v=\tfrac1y\); convert back at the end
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.In the substitution method, we first:
Explanation: Substitute one variable.
Q2.Solving $x+y=10$ and $x-y=4$ gives $x=$
Explanation: Add: $2x=14\Rightarrow x=7$.
Q3.The three algebraic methods are substitution, elimination and:
Explanation: Cross-multiplication.
Q4.If $2x=8$, then $x=$
Explanation: $x=4$.
Practice more MCQs → 📝 Assignment · 35 marks