Solving Linear Systems by Substitution Step by Step

A system of linear equations contains two or more equations that must be true at the same time. When a system has two variables, such as (x) and (y), the solution is the coordinate pair that satisfies both equations. On a graph, this solution is usually the point where two straight lines intersect.

Solving systems of linear equations by substitution is especially useful when one equation already has a variable by itself. The method turns a two-variable problem into a one-variable equation, making the algebra easier to manage and check.

Students in Australia may meet simultaneous equations in Years 9 and 10, senior secondary mathematics, or courses leading towards the HSC in New South Wales and the VCE in Victoria. Whether the problem involves dollars, distances, ticket prices, or coordinates, the same process applies.

Recognising a suitable system

Consider the system

[ y=2x+1 ]

[ 3x+y=11 ]

The first equation gives (y) directly in terms of (x). This makes it a good starting point for substitution. Since (y) is equal to (2x+1), that expression can replace (y) in the second equation.

A system can also be written in standard form, such as

[ 2x+y=7 ]

[ x-y=2 ]

Neither variable is isolated at first, but the second equation can quickly be rearranged:

[ x-y=2 ]

[ -y=2-x ]

[ y=x-2 ]

Once a variable has been isolated, the substitution method becomes straightforward.

Isolating one variable

The first main step is to rearrange one equation so that (x) or (y) stands alone. Choose the variable that requires the fewest operations. This reduces the chance of sign errors and keeps the calculations shorter.

For example:

[ 2x+y=9 ]

Subtract (2x) from both sides:

[ y=9-2x ]

Now use this form with the second equation:

[ x+3y=13 ]

The expression (9-2x) is equal to (y), so it can take (y)'s place in the second equation. The two equations have been linked without changing their meaning.

Be careful when rearranging negative terms. From (x-y=5), the correct expression for (y) is (y=x-5), not (y=5-x). Writing each algebraic operation on its own line is helpful, particularly when working through homework or preparing for an assessment.

Substituting into the other equation

Using the example above, substitute (y=9-2x) into (x+3y=13):

[ x+3(9-2x)=13 ]

The brackets show that the entire expression replaces (y). Expand and simplify:

[ x+27-6x=13 ]

[ -5x+27=13 ]

[ -5x=-14 ]

[ x=\frac{14}{5} ]

The two-variable system has now become a one-variable equation. This is the key advantage of substitution: one unknown is removed before the remaining value is found.

Next, replace (x) in either original rearranged equation:

[ y=9-2\left(\frac{14}{5}\right) ]

[ y=9-\frac{28}{5} ]

[ y=\frac{45}{5}-\frac{28}{5} ]

[ y=\frac{17}{5} ]

The solution is

[ \left(\frac{14}{5},\frac{17}{5}\right) ]

Decimals are also acceptable when the question requests them. In this case, the point is ((2.8,3.4)).

Working with whole-number answers

Some problems are designed to produce simpler values. For example:

[ y=x+4 ]

[ 2x+y=13 ]

Substitute (x+4) for (y):

[ 2x+(x+4)=13 ]

Simplify:

[ 3x+4=13 ]

[ 3x=9 ]

[ x=3 ]

Then use (y=x+4):

[ y=3+4=7 ]

The solution is ((3,7)). It is important to substitute the value back into an equation from the original system or a correctly rearranged version. This finds the second coordinate and completes the ordered pair.

When an equation contains brackets, distribute multiplication across every term inside them. For instance, (4(2x-3)) becomes (8x-12), not (8x-3). Combining like terms carefully helps prevent small slips from changing the final answer.

Checking the solution and interpreting graphs

A solution should satisfy both original equations. For ((3,7)), check the first equation:

[ y=x+4 ]

[ 7=3+4 ]

This is true. Check the second:

[ 2x+y=13 ]

[ 2(3)+7=13 ]

[ 13=13 ]

Both checks work, so the ordered pair is correct. Checking is especially valuable when negative numbers, fractions, or several brackets appear in the working.

Each linear equation represents a straight line. A system with one solution has lines that intersect once. Parallel lines have no solution because they never meet, while identical lines have infinitely many solutions because every point on one line lies on the other. Algebra can reveal these cases: a false statement such as (4=9) indicates no solution, while an identity such as (0=0) indicates infinitely many solutions.

Applying substitution to real problems

Suppose a school canteen sells sandwiches for (s) dollars and fruit cups for (f) dollars. A student buys two sandwiches and one fruit cup for $15, while another combination costs a different total. These relationships can be represented with simultaneous equations, then solved to find each price. In Australia, using dollar amounts and cents makes the situation familiar, whether the setting is a Brisbane school, a Melbourne shopping centre, or a community fundraiser.

Substitution also helps with distance and travel problems. A question might compare two cyclists travelling along a path near Adelaide, or describe fares on public transport in Sydney. The equations could represent total distance, time, or cost. The algebraic solution has meaning only after the variables are interpreted in context, so include units where appropriate.

Online maths tools and calculators can support the process, but they should not replace the algebraic reasoning. A useful routine is to rearrange one equation by hand, substitute it into the second, simplify line by line, and then use a calculator to check fractions or decimals. Step-by-step homework help is most useful when it explains why each operation is valid.

To remember the method, isolate one variable, substitute its expression into the other equation, solve for the remaining variable, and substitute back. Then check both original equations. The answer is the ordered pair that makes the entire system true.