Substitution method

The graphical method is visual but slow. The first algebraic shortcut is substitution: solve one equation for one variable in terms of the other, then plug into the second equation.

Algorithm

1. From one of the equations, express one variable (say $y$) in terms of the other (say $x$). Choose the equation that makes this step easiest , typically the one where some variable has a coefficient $\pm 1$.
2. Substitute this expression into the other equation. The result is a single linear equation in one variable.
3. Solve that equation to find one variable.
4. Substitute back to find the other variable.
5. Verify by plugging the pair into both original equations.

The method is mechanical and works whenever a unique solution exists. If you end up with a statement like "$0 = 0$", the system has infinitely many solutions; if you get "$0 = 5$" (a false statement), the system has no solution.

Why it works

Both equations are statements about the same $x$ and $y$. So if equation $(1)$ forces $y = f(x)$, then equation $(2)$ becomes a constraint purely in $x$ once we substitute. Solving that constraint pins down $x$, and substitution recovers $y$. The whole procedure is just clean algebra; the geometric meaning is "follow the first line until you hit the second".

It is worth seeing why substitution can fail to give a unique answer. If the two equations are dependent (one is a multiple of the other), the substituted equation will reduce to $0 = 0$ , meaning every $x$ on the line works. If the two equations are inconsistent (parallel, distinct lines), you get a contradiction like $0 = 7$.

Worked examples

Example 1. Solve $x + y = 14$ and $x - y = 4$.

From the second equation, $x = y + 4$. Substitute into the first: $(y + 4) + y = 14 \Rightarrow 2 y = 10 \Rightarrow y = 5$. So $x = 5 + 4 = 9$.

Solution $(x, y) = (9, 5)$. Verify: $9 + 5 = 14$ ✓, $9 - 5 = 4$ ✓.

Example 2. Solve $3 x + 2 y = 11$ and $2 x + 3 y = 4$.

Solve the first for $y$: $y = (11 - 3 x)/2$. Substitute: $2 x + 3 \cdot \frac{11 - 3 x}{2} = 4 \Rightarrow 4 x + 3(11 - 3 x) = 8 \Rightarrow 4 x + 33 - 9 x = 8 \Rightarrow -5 x = -25 \Rightarrow x = 5.$ Then $y = (11 - 15)/2 = -2$.

Solution $(5, -2)$.

Example 3. Solve $0.4 x + 0.3 y = 1.7$ and $0.7 x - 0.2 y = 0.8$.

Multiply each by $10$ to clear decimals: $4 x + 3 y = 17$ and $7 x - 2 y = 8$.

From the first, $y = (17 - 4 x)/3$. Substitute into the second: $7 x - 2 \cdot \frac{17 - 4 x}{3} = 8 \Rightarrow 21 x - 2(17 - 4 x) = 24 \Rightarrow 21 x - 34 + 8 x = 24 \Rightarrow 29 x = 58 \Rightarrow x = 2.$ Then $y = (17 - 8)/3 = 3$.

Solution $(2, 3)$.

Example 4. Solve $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = -2$.

Let $u = 1/x$ and $v = 1/y$. The system becomes $2 u + 3 v = 13$, $5 u - 4 v = -2$.

From the first, $u = (13 - 3 v)/2$. Substitute: $5 \cdot \frac{13 - 3 v}{2} - 4 v = -2 \Rightarrow 5(13 - 3 v) - 8 v = -4 \Rightarrow 65 - 15 v - 8 v = -4 \Rightarrow -23 v = -69 \Rightarrow v = 3.$ Then $u = (13 - 9)/2 = 2$. So $x = 1/u = 1/2$ and $y = 1/v = 1/3$.

Example 5. Solve $x - y = 3$ and $2 x - 2 y = 9$.

From the first, $x = y + 3$. Substitute: $2(y + 3) - 2 y = 9 \Rightarrow 6 = 9$. Contradiction. No solution.

(Ratio check: $1/2, -1/-2, -3/-9$. First two equal $1/2$; third equals $1/3$. Indeed inconsistent.)

Try it yourself

1. Solve by substitution: $x + 2 y = -1, 2 x - 3 y = 12$.
2. Solve: $0.2 x + 0.3 y = 1.3, 0.4 x + 0.5 y = 2.3$.
3. Solve: $\sqrt{2} x + \sqrt{3} y = 0, \sqrt{3} x - \sqrt{8} y = 0$.
4. Solve: $\dfrac{3}{x} + \dfrac{2}{y} = 12, \dfrac{2}{x} + \dfrac{3}{y} = 13$.
5. Solve: $x + y = 5, 3 x + 3 y = 15$. (Decide first by inspection.)
6. Solve: $x - 4 y - 14 = 0, 5 x - y - 13 = 0$.
7. Solve: $2 x - 3 y = -8, 4 x + y = 9$.
8. Solve: $\dfrac{x}{2} + \dfrac{2 y}{3} = -1, x - \dfrac{y}{3} = 3$.
9. Solve: $3(x - 1) + 2(y - 2) = 5, 2(x + 1) - 5(y + 2) = 0$.
10. Show by substitution that $x + 2 y = 5, 2 x + 4 y = 10$ has infinitely many solutions; describe them all.

Pitfalls / Insight

• Always choose the easiest variable to isolate. Coefficients of $\pm 1$ make life simple.
• Clear fractions and decimals first by multiplying through , substitution gets messy with denominators.
• Reach for a substitution like $u = 1/x, v = 1/y$ when reciprocals appear. The new system is linear and standard.

Insight. Substitution is the most "human" of the algebraic methods , it just chains two equations together. When the structure is friendly, no other method is faster.