Substitution method

Substitution is a fixed sequence, not a bag of tricks — the same five moves solve every pair of linear equations you will meet in this chapter.

Choose wisely. Scan both equations for a variable whose coefficient is already $1$ or $-1$; isolating that one avoids fractions for as long as possible. If every coefficient is something else, elimination (covered separately) is often faster, but substitution still works.

Isolate, then substitute. Write the chosen variable as an expression in the other, and replace every occurrence of it in the other equation — never the one you took it from, or you just get an identity like $0=0$.

Solve one variable, then find the other. The substitution leaves a single linear equation in one unknown. Solve it, then back-substitute into the isolated expression to recover the second value, and check both numbers in the original pair. In general, for $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ with $b_1 \neq 0$, isolating $y$ from the first equation gives $y = \dfrac{c_1-a_1x}{b_1}$, and substituting into the second collapses the system to

where: $x, y$ are the unknowns, $a_1,b_1,c_1$ are the coefficients and constant of the first equation, and $a_2,b_2,c_2$ those of the second. Solving this single equation for $x$, then back-substituting, gives the complete solution — provided the lines are not parallel or identical (that special case is flagged in Common traps below).

Applies directly to word problems. Ages, digits, fractions, speed-distance-time and cost questions all reduce to a pair of linear equations once you name the unknowns, and substitution is usually the natural way to solve them because the sentence itself often hands you one variable "in terms of" the other. For instance, the sum of two numbers is 58 and their difference is 12 becomes $x+y=58$ and $x-y=12$; isolating $x=y+12$ from the second and substituting into the first gives $2y+12=58$, so $y=23$ and $x=35$ — the same isolate-substitute-solve-back-substitute-verify sequence, just with a sentence to translate first instead of an equation handed to you ready-made.

Solve using substitution: $2x + 3y = 11$ and $x - 2y = -7$.

Step 1. Isolate $x$ from the second equation, since its coefficient is $1$: $x = 2y - 7$.

Step 2. Substitute into the first equation: $2(2y-7) + 3y = 11$.

Step 3. Solve for $y$: $4y - 14 + 3y = 11 \implies 7y = 25 \implies y = \dfrac{25}{7}$.

Step 4. Back-substitute: $x = 2\left(\dfrac{25}{7}\right) - 7 = \dfrac{50}{7} - \dfrac{49}{7} = \dfrac{1}{7}$.

Step 5. Verify: $2\left(\dfrac17\right) + 3\left(\dfrac{25}{7}\right) = \dfrac{2}{7}+\dfrac{75}{7} = \dfrac{77}{7} = 11$ ✓, and $\dfrac17 - 2\left(\dfrac{25}{7}\right) = \dfrac17 - \dfrac{50}{7} = -\dfrac{49}{7} = -7$ ✓. So $x = \dfrac17$, $y = \dfrac{25}{7}$ — a valid solution even though both values are fractions.

Competitive stretch

The same pair can be checked against the general formula from the derivation above. Here $a_1=2,\ b_1=3,\ c_1=11,\ a_2=1,\ b_2=-2,\ c_2=-7$. Then $x = \dfrac{c_2b_1-b_2c_1}{a_2b_1-a_1b_2} = \dfrac{(-7)(3)-(-2)(11)}{(1)(3)-(2)(-2)} = \dfrac{-21+22}{3+4} = \dfrac{1}{7}$, matching the worked value exactly — a fast way to sanity-check an answer on a calculator-free paper without repeating every step.

• Substituting into the same equation the expression was taken from, instead of the other one — this only produces an identity like $0=0$ and gives no new information.
• Sign errors when distributing a negative coefficient across brackets, e.g. writing $2(2y-7)$ as $4y-7$ instead of the correct $4y-14$, or forgetting to back-substitute and leaving the answer as only one variable's value.
• Rounding fractional answers too early. A valid solution can be a fraction like $\dfrac17$; keep exact fractions through every step, since early rounding can make the final verification fail.
• Assuming a unique solution always exists. If $a_2b_1 - a_1b_2 = 0$ (parallel or identical lines), substitution collapses to $0 = 0$ (infinitely many solutions) or a false statement like $0 = 5$ (no solution) — recognise this rather than hunting for an error, and in word problems, always name the unknowns and write both equations before substituting.

Ask the tutor: if your substitution keeps producing messy fractions or an identity like 0 equals 0, share the exact pair of equations and the AI tutor will help you spot where to adjust your approach.