Systems of Equations

Every method for solving a system does the same thing: reduce two equations in two unknowns to one equation in one unknown. The methods differ only in how they get rid of the first variable.

Elimination

Add or subtract the equations so one variable cancels.

4x + 3y = 17 2x - 3y = -5

The y terms are already opposites, so add the equations straight down:

• 4x plus 2x is 6x.
• 3y plus -3y is 0.
• 17 plus -5 is 12.

So 6x = 12, and x = 2.

Put that back into the second equation: 4 - 3y = -5, so -3y = -9 and y = 3.

Now the other case, where the coefficients match rather than oppose:

2x + 3y = 14 2x - y = 6

The x coefficients are both positive 2, so subtract the second from the first:

• 2x minus 2x is 0.
• 3y minus -y is 4y.
• 14 minus 6 is 8.

So 4y = 8, and y = 2. Then 2x - 2 = 6, so x = 4.

Matching signs subtract. Opposite signs add. Getting that backward is the most common error in the method, and it produces a system that has not been reduced at all: adding those last two equations gives 4x + 2y = 20, which still has two unknowns in it.

When nothing matches

Multiply one equation until something does.

3x + 4y = 17 5x - 2y = 11

Nothing cancels as written. Doubling the second equation gives 10x - 4y = 22, and now the y terms are opposites. Add:

• 3x plus 10x is 13x.
• 4y plus -4y is 0.
• 17 plus 22 is 39.

So 13x = 39 and x = 3. Then 5(3) - 2y = 11, so -2y = -4 and y = 2.

You choose what to multiply by, so choose the target that keeps the arithmetic small. Here doubling one equation was enough because 2 divides 4. Had you aimed at x instead, you would have needed to multiply the first by 5 and the second by 3, which is more work for the same answer.

Substitution

Solve one equation for one variable, then put that into the other.

Use it when a variable is already alone, or has a coefficient of 1.

y = 2x - 5 3x + y = 15

The first equation already gives y. Substitute it into the second:

3x + (2x - 5) = 15

• Combine: 5x - 5 = 15.
• Add 5: 5x = 20.
• Divide: x = 4.

Back-substitute into the simplest equation: y = 2(4) - 5 = 3.

Use parentheses when substituting. Dropping them loses the sign on the -5 and produces 3x + 2x - 5 correctly here, but with a subtraction in front it would not.

Making a variable alone

If neither variable is isolated, isolate the one with a coefficient of 1.

x + 3y = 14 2x - y = 7

The x in the first equation has coefficient 1, so x = 14 - 3y. Substitute into the second:

2(14 - 3y) - y = 7

• Distribute: 28 - 6y - y = 7.
• Combine: 28 - 7y = 7.
• Subtract 28: -7y = -21.
• Divide: y = 3.

Then x = 14 - 9 = 5.

Check in the equation you did not use last: 2 times 5 minus 3 is 7. Correct.

Choosing a method in five seconds

Both methods always work. The table is about speed, not legality.

Word problems

The hard part is writing the two equations, and the pattern is stable: one equation counts things, the other totals values.

Tickets cost $8 for adults and $5 for children. 90 tickets sold for $606. How many adults?

• Count: a + c = 90.
• Value: 8a + 5c = 606.

From the first, c = 90 - a. Substitute:

8a + 5(90 - a) = 606

• Distribute: 8a + 450 - 5a = 606.
• Combine: 3a + 450 = 606.
• Subtract: 3a = 156.
• Divide: a = 52.

So 52 adults and 38 children. Check: 8 times 52 is 416, 5 times 38 is 190, and 416 plus 190 is 606. Correct.

No solution, or infinitely many

Enlarge

Zoom in

Most systems have exactly one answer - one point where the two lines cross. Two cases do not:

• No solution: the variables cancel and leave something false, like 0 = 1. The two lines are parallel - same slope, different intercepts - and never meet. 2x + 6y = 9 and x + 3y = 5 is this case: double the second and it says 2x + 6y = 10.
• Infinitely many: everything cancels and leaves something always true, like 0 = 0. The two equations are the same line: 9x - 12y = 36 is just 3x - 4y = 12 multiplied by 3.

The same thing happens with a single equation: 4(x - 3) = 2(2x + 5) becomes -12 = 10, so there is no solution.

What you can skip

Across the 20 questions on this topic:

• Three equations in three unknowns and matrix methods never appear.
• Solving by graphing. No question asks you to draw the lines; elimination and substitution cover every system.

Where people lose points

Adding when the signs match, or subtracting when they are opposite.

Distributing a subtraction wrongly when subtracting one equation from another. Every term on that side changes sign.

Stopping after one variable. If the question asks for y and you found x first, you are not finished. Worse, if it asks for x + y, both are needed and each alone is a choice.

Substituting without parentheses and losing a sign.

Checking in the equation you just used, which will always work even if you made an error earlier. Check in the other one.

Work one in under a minute

4x + y = 23 x - y = 2

Find x + y.

The y terms are already opposites, so add:

• 4x plus x is 5x.
• y plus -y is 0.
• 23 plus 2 is 25.

So 5x = 25 and x = 5. Then from the second equation, 5 - y = 2, so y = 3.

The question asked for x + y, which is 8.

Both 5 and 3 will be choices.

Where this leads

Systems are the algebra behind mixture, coin and ticket word problems, and the same elimination move reappears whenever two conditions constrain two quantities.