learn › Algebra

Systems of equations

lesson · about 3 minutes

Two equations, two unknowns, one answer.

Pen and paper is fine · no calculator needed why?

Opens at level 24. You're level 1. You can read and practice here now.

the idea

A system is two equations that share two unknowns, x and y. The answer is the one pair of values that makes both equations true at once.

The plan is always the same: get rid of one letter, solve for the other, then put that value back. Elimination adds or subtracts the equations. Substitution swaps y for what it equals. Give x first, then y.

techniques

Add or subtract to eliminate

One letter has the same or opposite numbers in front of it in both equations.

  1. Opposite numbers in front, like y and −y: add the equations.
  2. The same number in front: subtract the second equation from the first.
  3. Solve for the letter that is left.
  4. Put it into the first equation to find the other letter.
worked example

Example: Solve the system 2x + y = 11 and 3x − y = 4. What is y?

  1. The y terms are opposites, so add: 5x = 15.
  2. So x = 3.
  3. Put x = 3 into the first: 6 + y = 11, so y = 5.

Answer: 5

Substitute for y

One equation already says y = something.

  1. Replace y in the other equation with what it equals, in brackets.
  2. Multiply out the bracket, every term inside it.
  3. Solve for x, then put x back into the y equation to get y.
worked example

Example: Solve the system y = x + 2 and 3x + 2y = 19. What is x?

  1. Substitute: 3x + 2(x + 2) = 19.
  2. Multiply out: 3x + 2x + 4 = 19, so 5x + 4 = 19.
  3. 5x = 15, so x = 3.

Answer: 3

Two items, one total

Two prices, a total count, and a total amount of money.

  1. Let a letter count the pricier item. The cheaper count is the total count minus it.
  2. Write one equation for the money, and multiply out the bracket.
  3. Solve. For the cheaper item, subtract from the total count.
worked example

Example: A bakery sells pies for $9 and tarts for $4. It sold 15 items for $85. How many tarts did it sell?

  1. Let p be pies, so 15 − p are tarts.
  2. 9p + 4(15 − p) = 85, so 5p + 60 = 85.
  3. So 5p = 25 and p = 5.
  4. Tarts: 15 − 5 = 10.

Answer: 10

watch out for

practice

Sign in to try one

Solve by elimination

worked example

Solve the system: −x − 3y = −6 −x − 4y = −8 Enter x, y.

Answer: (0, 2)

  1. Subtract the second equation from the first: y = 2.
  2. Put y = 2 into the first equation: −x = −6 − (−6) = 0, so x = 0.

Solve by substitution

worked example

Solve the system: y = 2x + 4 3x + 5y = −19 Enter x, y.

Answer: (-3, -2)

  1. Substitute: 3x + 5(2x + 4) = −19.
  2. So 13x + 20 = −19, which gives 13x = −39.
  3. x = −3, and y = 2 × (−3) + 4 = −2.

Two-item word problem

worked example

A café sells large coffees for $5 and small coffees for $4. One morning it sold 58 coffees for $276. How many large coffees did it sell?

Answer: 44

  1. Let l be the number of large coffees, so 58 − l are small coffees: 5l + 4(58 − l) = 276.
  2. That gives l + 232 = 276, so l = 276 − 232 = 44.

Sign in to start