Solving Linear Systems By Substitution Examples

Solving Linear Systems by SubstitutionExamples

The substitution method is one of the most straightforward techniques for finding the solution of a system of linear equations. Worth adding: by isolating one variable in a single equation and then plugging that expression into the other equation, the problem is reduced to a single‑variable equation that can be solved with basic algebra. This approach works especially well when one of the equations already has a coefficient of 1 or − 1 for a variable, making isolation quick and error‑free. Below we walk through the core steps, illustrate the method with several worked‑out examples, explain why the procedure is mathematically sound, and answer common questions that learners often encounter.

How the Substitution Method Works

Before diving into examples, it helps to outline the general procedure. Following these steps consistently reduces the chance of algebraic slips and keeps the solution process transparent.

Choose an equation and solve for one variable.
Pick either equation and isolate x or y (whichever leads to simpler arithmetic). 2. Substitute the expression into the other equation.
Replace the chosen variable in the second equation with the expression obtained in step 1.
Solve the resulting single‑variable equation.
Use addition, subtraction, multiplication, or division to find the numeric value of the remaining variable. 4. Back‑substitute to find the other variable.
Plug the value from step 3 into the expression from step 1 to compute the second variable.
Check the solution in both original equations.
Verify that the ordered pair satisfies each equation; this catches any arithmetic mistakes.

When the system has infinitely many solutions or no solution, the substitution process will reveal that as well (e.g., you may end up with a true statement like 0 = 0 or a false one like 5 = 0).

Worked‑Out Examples

Example 1: Simple Integer Coefficients

Solve the system
[ \begin{cases} 2x + y = 7 \ x - 3y = -5 \end{cases} ]

Step 1 – Isolate a variable.
From the first equation, solve for y: [ y = 7 - 2x]

Step 2 – Substitute.
Replace y in the second equation:
[ x - 3(7 - 2x) = -5 ]

Step 3 – Solve for x.
[ \begin{aligned} x - 21 + 6x &= -5 \ 7x - 21 &= -5 \ 7x &= 16 \ x &= \frac{16}{7} \end{aligned} ]

Step 4 – Back‑substitute for y. [ y = 7 - 2\left(\frac{16}{7}\right) = 7 - \frac{32}{7} = \frac{49}{7} - \frac{32}{7} = \frac{17}{7} ]

Step 5 – Check.
Plug ((x,y)=\left(\frac{16}{7},\frac{17}{7}\right)) into both original equations; both hold true.

Solution: (\displaystyle \left(\frac{16}{7},\frac{17}{7}\right)).

Example 2: One Equation Already Solved for a Variable

Solve the system
[\begin{cases} y = 4x - 1 \ 3x + 2y = 10 \end{cases} ]

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Because the first equation already expresses y in terms of x, we can skip the isolation step.

Step 2 – Substitute.
[ 3x + 2(4x - 1) = 10 ]

Step 3 – Solve for x. [ \begin{aligned} 3x + 8x - 2 &= 10 \ 11x - 2 &= 10 \ 11x &= 12 \ x &= \frac{12}{11} \end{aligned} ]

Step 4 – Back‑substitute for y.
[ y = 4\left(\frac{12}{11}\right) - 1 = \frac{48}{11} - \frac{11}{11} = \frac{37}{11} ]

Step 5 – Check.
Both equations are satisfied.

Solution: (\displaystyle \left(\frac{12}{11},\frac{37}{11}\right)).

Example 3: Fractions and Decimals

Solve the system
[ \begin{cases} \frac{1}{2}x - \frac{1}{3}y = 1 \ 0.4x + 0.2y = 2\end{cases} ]

It is often helpful to clear fractions or decimals first, but we can also substitute directly.

Step 1 – Isolate x from the first equation.
Multiply both sides by 6 to eliminate denominators:
[ 3x - 2y = 6 \quad\Rightarrow\quad 3x = 6 + 2y \quad\Rightarrow\quad x = \frac{6 + 2y}{3} ]

Step 2 – Substitute into the second equation.
[ 0.4\left(\frac{6 + 2y}{3}\right) + 0.2y = 2 ]

Step 3 – Solve for y.
First, compute (0.4/3 = \frac{0.4}{3} = \frac{2}{15}) (since (0.4 = \frac{2}{5})).
[ \frac{2}{15}(6 + 2y) + 0.2y = 2 ] [ \frac{12}{15} + \frac{4}{15}y + 0.2y = 2 ] Convert (0.2y) to fifteenths: (0.2 = \frac{3}{15}).
[ \frac{12}{15} + \frac{4}{15}y + \frac{3}{15}y = 2 ] [ \frac{12}{15} + \frac{7}{15}y = 2 ] Multiply by 15:
[ 12 + 7y = 30 \quad\Rightarrow\quad 7y = 18 \quad\Rightarrow\quad y = \frac{18}{7} ]

Step 4 – Back‑substitute for x.
[ x = \frac{6 + 2\left(\frac{18}{7}\right)}{3} = \frac{6 + \frac{36}{7}}{3} = \frac{\frac{42}{7} + \frac{36}{7}}{3} = \frac{\frac{78}{7}}{3} = \frac{7

}{78/7} = \frac{78}{21} = \frac{26}{7} ]

Step 5 – Check.
Substitute (x=\frac{26}{7}) and (y=\frac{18}{7}) into both original equations; both are satisfied.

Solution: (\displaystyle \left(\frac{26}{7},\frac{18}{7}\right)).

Summary

The substitution method works by isolating one variable in one equation, substituting that expression into the other equation, and solving the resulting single-variable equation. But it is especially efficient when one equation is already solved for a variable or can be easily rearranged. The method handles systems with integers, fractions, or decimals, though clearing fractions or decimals first can simplify arithmetic. Always verify the final ordered pair in both original equations to confirm correctness.