In short

The substitution method turns a two-variable problem into a one-variable problem in two clean strokes. Step 1 — isolate: pick one equation, isolate one variable (say y), so it sits alone on the left as an expression in x. Step 2 — inject: wherever y appears in the other equation, replace it by that expression. The other equation now has only x in it. Solve for x, then back-substitute into the isolated form to get y. It is one of the three standard methods for CBSE Class 10 (the other two being graphical and elimination), and it works because replacing equals with equals is always legal.

You are staring at two equations:

x + y = 7, \qquad 2x - y = 5.

Each one alone is an entire line — infinitely many (x,y) pairs satisfy it. But the crossing point, the single (x, y) that satisfies both at once, is what you want. The substitution method finds it by collapsing the system into a single equation in a single variable. Here is the trick, in two moves.

The two moves

Move 1 — isolate. Take one equation and rearrange it so a chosen variable sits alone on one side. From x + y = 7, isolate y: \,y = 7 - x. Why: subtracting x from both sides preserves equality — both lines, before and after, have the same set of solutions.

Move 2 — inject. The expression 7 - x is now a certified replacement for y — anywhere y appears, you can write 7 - x instead, and the truth value will not change. So in the other equation, 2x - y = 5, replace y by 7 - x:

2x - (7 - x) = 5.

This new equation has only x in it. Why: the moment you injected 7 - x in place of y, every reference to y vanished from the second equation. One variable left, and one equation — exactly enough to solve.

Solve: 2x - 7 + x = 5 \implies 3x = 12 \implies x = 4. Back-substitute into y = 7 - x to get y = 3. The crossing point is (4, 3). Done.

Watch the variable collapse

Click the buttons. Equation 1 isolates y, then the isolated expression slides over and replaces the y in Equation 2.

Equation 1: x + y = 7 Equation 2: 2x − y = 5 Isolated: y = 7 − x Collapsed: 2x − (7 − x) = 5

Press Step 1 to isolate $y$ from equation 1.

The flow, drawn

Step-by-step substitution flow Two equations on top. An arrow labelled "isolate y" points down-left to the isolated form y = 7 - x. A second arrow labelled "inject" points from the isolated form into the second equation, producing a single-variable equation 2x - (7 - x) = 5 at the bottom. x + y = 7 2x − y = 5 isolate y y = 7 − x inject (7 − x) for y 2x − (7 − x) = 5 → one variable
Two equations in, one equation out. The variable $y$ has been collapsed away.

Worked examples

Example 1 — the clean case

Solve the system \{\,x + y = 7,\ 2x - y = 5\,\}.

Isolate y in equation 1: y = 7 - x.

Inject into equation 2:

2x - (7 - x) = 5 \implies 2x - 7 + x = 5 \implies 3x = 12 \implies x = 4.

Back-substitute: y = 7 - 4 = 3.

Solution: (x, y) = (4, 3). Check: 4 + 3 = 7 and 2(4) - 3 = 5. Both satisfied.

Example 2 — already isolated

Solve \{\,2x + 3y = 13,\ x = y + 1\,\}.

The second equation already has x isolated — no work needed in step 1. Inject x = y + 1 into equation 1:

2(y + 1) + 3y = 13 \implies 2y + 2 + 3y = 13 \implies 5y = 11 \implies y = \tfrac{11}{5}.

Back-substitute: x = y + 1 = \tfrac{11}{5} + 1 = \tfrac{16}{5}.

Solution: \left(\tfrac{16}{5},\ \tfrac{11}{5}\right). Why this case is fast: when one equation is already in x = \ldots or y = \ldots form, you skip the isolation step and go straight to injection.

Example 3 — when substitution is messy

Solve \{\,3x + 4y = 11,\ 2x - 5y = 8\,\}.

Neither variable isolates cleanly — every isolation produces a fraction. Push through anyway. From equation 1, isolate x:

x = \frac{11 - 4y}{3}.

Inject into equation 2:

2 \cdot \frac{11 - 4y}{3} - 5y = 8.

Multiply through by 3 to clear the denominator:

2(11 - 4y) - 15y = 24 \implies 22 - 8y - 15y = 24 \implies -23y = 2 \implies y = -\tfrac{2}{23}.

Then x = \dfrac{11 - 4(-2/23)}{3} = \dfrac{11 + 8/23}{3} = \dfrac{(253 + 8)/23}{3} = \dfrac{261}{69} = \dfrac{87}{23}.

Solution: \left(\tfrac{87}{23},\ -\tfrac{2}{23}\right). It works — but the arithmetic is ugly. For a system like this, elimination (multiply equation 1 by 5 and equation 2 by 4, then add to kill y) is usually faster. Substitution is at its best when at least one variable is easy to isolate; otherwise consider switching tools.

Why this is allowed at all

The whole method rests on one principle: if A = B, then anywhere you see A, you may write B instead, and any true statement stays true. Why: that is literally what the equals sign means — A and B are the same number. Replacing equals by equals never invents or destroys solutions.

So when you say "y equals 7 - x" and then write 7 - x in place of y, you are not approximating, not guessing — you are using the certificate that the first equation hands you. Every (x, y) that satisfies both originals also satisfies the collapsed equation, and every x that solves the collapsed equation pairs (via back-substitution) with exactly one y that satisfies both originals.

When to reach for substitution

References

  1. NCERT Class 10 Mathematics, Chapter 3: Pair of Linear Equations in Two Variables — the official Indian syllabus treatment.
  2. Substitution method (Wikipedia) — formal statement and worked examples.
  3. Khan Academy: Solving systems by substitution — additional practice problems.