Systems of linear equations

What "solve the system" means#

A solution of a system is a pair of numbers that makes both equations true at the same time. Both, not one: a pair that fits only the first equation is not a solution. That is why the answer is always written as a pair, for example $(5, 3)$, and checked by substituting into both lines.

The two standard methods differ only in how they get rid of the second variable. Substitution is handy when one of the equations already has a variable on its own, or it is easy to isolate. Elimination is handy when the coefficients of one variable are opposites, or are easy to make opposites.

Elimination: a simple case#

The system: $x + y = 10$ and $x - y = 4$.

Step 1. Look at the coefficients of $y$: $+1$ in the first equation, $-1$ in the second. They are opposites, so $y$ will drop out when you add.

Step 2. Add the left sides and the right sides separately: $(x + y) + (x - y) = 10 + 4$, that is $2x = 14$.

Step 3. Divide by 2: $x = 7$.

Step 4. Go back to either original equation — the first is simpler: $7 + y = 10$, so $y = 3$.

Step 5. Check in the second equation: $7 - 3 = 4$. Correct. The answer is $(7, 3)$.

Step 4 is the one people skip most: having found $x$, they think they are done. But the answer to a system is a pair; a single number is not an answer.

The substitution method#

The system: $x + 2y = 11$ and $3x - y = 12$.

Step 1. Choose what to isolate. $x$ from the first equation is easiest: its coefficient is one, so no fractions appear.

Step 2. Isolate it: $x = 11 - 2y$.

Step 3. Substitute this expression for $x$ in the second equation — always in brackets: $3(11 - 2y) - y = 12$.

Step 4. Expand: $33 - 6y - y = 12$. The 3 multiplies both terms, and this is where a minus sign gets lost.

Step 5. Combine like terms and move the number: $-7y = 12 - 33$, that is $-7y = -21$.

Step 6. Divide by $-7$: $y = 3$.

Step 7. Put it into the expression from step 2: $x = 11 - 2 \cdot 3 = 5$.

Step 8. Check in the second equation: $3 \cdot 5 - 3 = 12$. The answer is $(5, 3)$.

Elimination: a harder example#

The system: $2x + 3y = 16$ and $5x - 3y = 19$.

Step 1. The coefficients of $y$ are $+3$ and $-3$ — opposites, so you can add the equations straight away without multiplying anything.

Step 2. Add: $7x = 35$.

Step 3. So $x = 5$.

Step 4. Substitute into the first equation: $2 \cdot 5 + 3y = 16$, so $3y = 6$ and $y = 2$.

Step 5. Check in the second: $25 - 6 = 19$. The answer is $(5, 2)$.

If there were no opposite coefficients — say $2x$ and $3x$ — you would first multiply the equations term by term: the first by 3, the second by $-2$, which gives $6x$ and $-6x$. Multiply the whole line, including the right side.

Where systems come from in word problems#

A typical problem: a rectangle has a perimeter of 20 cm, and one side is 2 cm longer than the other. Call the sides $a$ and $b$. The first equation is about the perimeter: $2(a + b) = 20$, that is $a + b = 10$. The second is about the difference: $a = b + 2$. Substitute: $b + 2 + b = 10$, so $2b = 8$, $b = 4$ and $a = 6$. Check: the perimeter is $2 \cdot (6+4) = 20$ and the difference is 2 cm — it fits. The formulas behind the first equation are covered in area and perimeter.

In problems like this the mistake is usually not in solving the system but in setting it up: the variables are never described in words. Write the meaning next to them — "$a$ is the length in centimetres" — and check that each equation reads as a sentence from the problem.

Special cases and mistakes#

If eliminating a variable leaves you with $0 = 5$, there is no solution: the lines are parallel. If you get $0 = 0$, there are infinitely many solutions: both equations describe the same line. This is not an arithmetic error but a legitimate answer.

Adding the left sides and forgetting the right sides. You have to add both sides, or the equation breaks.

Substituting the value in the wrong place: after substituting into the second equation, people get an identity and start solving it all over again. Go back to the original equation, or to the expression you got when you isolated the variable.

Skipping the check. Substituting the pair into both lines takes one line and gives full confidence in the answer. The moves inside each line are the same as in how to solve linear equations.

Step-by-step plan

1. Step 1 — isolating a variableFrom equations like x + 2y = 11, isolate x and then y; aim for forms without fractions.
2. Step 2 — substitutionFive systems by substitution, always putting the substituted expression in brackets.
3. Step 3 — eliminationFive systems with ready-made opposite coefficients, no multiplying needed.
4. Step 4 — multiplying linesSystems where you must match the coefficients first: multiply the whole line, right side included.
5. Step 5 — word problemsSet up systems about perimeters, ages and purchases; describe the variables in words and check the answer against the problem.

Check yourself

Sources

• Khan Academy — Algebra 1The systems of equations unit, with step-by-step hints
  free
• OpenStax Elementary Algebra 2eFree textbook; chapter 5 covers graphing, substitution and elimination
  free

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