Many scientific problems lead to simultaneous equations containing quantities which need to be calculated. The simplest case is two simultaneous equations in two unknowns, say x and y.

**Example 1:** To start to see how we can solve such relations, consider:

4x + y |
= | 9 |

3x |
= | 6 |

There are two unknown variables x and y. However, the bottom equation only involves x and is solved by x = 2. We can then substitute this into the top equation to find:

4 × 2 + y |
= | 9 |

y |
= | 9 − 8 |

y |
= | 1 |

The full solution is therefore *x* = 2, *y* = 1.

**Exercise 1:** Solve the following pairs of simultaneous equations:

(a)*x* + *y* = 3 and *x* = 2

**Solution:** Substituting x = 2 into x + y = 3 we obtain:

2 + y |
= | 3 |

y |
= | 3 − 2 |

y |
= | 1 |

The solution is thus: x = 2, y = 1.

(b)4*x* − *y* = 10 and *y* = 2

**Solution:** Substituting y = 2 into 4x − y = 10 yields:

4x − 2 |
= | 10 |

4x |
= | 12 |

x |
= | 3 |

The solution is thus: x = 3, y = 2.

(c)*z* − *x* = 2 and 2*x* = −2

**Solution:** From: 2x = −2 we have that x = −1. Inserting this into z − x = 2 we find:

z − (−1) |
= | 2 |

z + 1 |
= | 2 |

z |
= | 1 |

The solution is thus: x = −1, z = 1.

(d)3*t* + 2*s* = 0 and *s* + 1 = 0

**Solution:** From: s + 1 = 2, we have: s = 1 and this can be inserted into: 3t + 2s to give:

3t + 2 |
= | 0 |

3t |
= | −2 |

t |
= | −23 |

The solution is thus: s = 1, t = ^{2}⁄_{3}.

Click on questions to reveal solutions

**Quiz 1:** What value of y solves the pair of equations: *x* + 2*y* = 0 and *x* = −2?

(a)12Incorrect - please try again!

(b)4Incorrect - please try again!

(c)8Incorrect - please try again!

(d)6Correct - well done!

**Explanation:** Substituting x = −2 into x + 2y = 10 yields:

−2 + 2y |
= | 10 |

2y |
= | 12 |

y |
= | 6 |

The solution is thus: x = −2, y = 6.

More generally both equations may involve both unknowns.

**Example 2:** Consider:

x + y = 4 |
(1) |
---|---|

x − y = 2 |
(2) |

Now add the left hand side of (1) to the left hand side of (2) and the right hand side of (1) to the right hand side of (2). The y’s cancel and we get an equation for x alone:

x + y + x − y |
= | 4 + 2 |

2x |
= | 6 |

which implies that x = 3. We can now insert this into (1) and so obtain:

3 + *y* = 4, ⇒ *y* = 4 − 3 = 1

In other words the full solution is x = 3 , y = 1

It is easy to check that you have the correct solution to simultaneous equations by substituting your answers back into the original equations.
We have already used (1) to find y, so let’s check that (2) is correctly solved. We get *x* − *y* = 3 − 1 = 2

Always make such a check!

**Exercise 2:** Solve the following pairs of equations:

(a)*x* + *y* = 5 and *x* − *y* = 1

**Solution:** Adding these equations yields:

2x |
= | 6 |

so x = 3. This can now be inserted into the first equation to give:

3 + y |
= | 5 |

y |
= | 2 |

These results can be checked by inserting them into the second equation:

(b)4*x* + 3*y* = 7 and *x* − 3*y* = −2

**Solution:** Adding these equations yields:

4x + 3y + x − 3y |
= | 7 − 2 |

5x |
= | 5 |

x |
= | 1 |

Substituting x = 1 into the first equation yields:

4 + 3y |
= | 7 |

3y |
= | 3 |

y |
= | 1 |

This can now be checked by substitution into:

**Example 3:** Consider:

x + y = 4 |
(1) |
---|---|

x − y = 2 |
(2) |

Now add the left hand side of (1) to the left hand side of (2) and the right hand side of (1) to the right hand side of (2). The ys cancel and we get an equation for x alone:

x + y + x − y |
= | 4 + 2 |

2x |
= | 6 |

which implies that x = 3. We can now insert this into (1) and so obtain:

3 + *y* = 4, ⇒ *y* = 4 − 3 = 1

In other words the full solution is x = 3, y = 1

**Quiz 2:** Solve the following simultaneous equations and select the correct result:

3x + 3y = 0 |
(1) |
---|---|

2x + 3y = 1 |
(2) |

(a)*x* = 0, *y* = 0Incorrect - please try again!

(b)*x* = -1, *y* = 1Correct - well done!

(c)*x* = 0, *y* = 1Incorrect - please try again!

(d)*x* = 3, *y* = 2Incorrect - please try again!

**Explanation:** Subtracting (1) from (2) yields:

3x + 3y − (2x + 3y) |
= | 0 |

x |
= | -1 |

This can now be substituted into (1) to yield:

−3 + 3y |
= | 0 |

3y |
= | 3 |

y |
= | 1 |

Check the solution, *x* = -1, *y* = 1, by substitution into (2)

The first step in solving a system of two simultaneous equations is to eliminate one of the variables. This can be done by making the coefficient of x the same in each equation.

**Example 4:** Consider:

3x + 2y = 4 |
(1) |
---|---|

2x + y = 3 |
(2) |

If we multiply (1) by 2 and (2) by 3, we get:

6x + 4y |
= | 8 |

6x + 3y |
= | 9 |

We see that the coefficient of x is now the same in each equation! Subtracting them cancels (‘eliminates’) x and we can solve the simultaneous equations using the methods described above.

**Example 5:** We now work through an example. Consider:

5x + 3y = 7 |
(1) |
---|---|

4x + 5y = 3 |
(2) |

Multiply (1) by 4 (which is the coefficient of x in (2)) and also multiply (2) by 5 (the coefficient of x in (1)):

20x + 12y = 28 |
(3) |
---|---|

20x + 25y = 15 |
(4) |

The coefficient of x is now the same in both equations. Subtracting (4)−(3) eliminates x:

25*y* − 12*y* = 15 − 28 , ⇒ 13*y* = −13

i.e., we have y = −1. Substituting this into (1) gives:

5*x* − 3 = 7 , ⇒ 5*x* = 10

so that x = 2. Now check that x = 2, y = −1 by substitution into (2)!

**Quiz 3:** To eliminate x from the following simultaneous equations,
what should you multiply them by?

3x − 2y = 7 |
(1) |
---|---|

4x − 5y = 7 |
(2) |

(a)7 & 7Incorrect - please try again!

(b)4 & 3Correct - well done!

(c)3 & −2Incorrect - please try again!

(d)3 & 4Incorrect - please try again!

**Explanation:** To
eliminate x we have to multiply (1) by 4 and (2) by 3. This yields:

12x − 8y = 28 |
(3) |
---|---|

12x − 15y = 21 |
(4) |

The x coefficient is then the same in each equation and so subtracting (4) from (3) indeed eliminates x.

**Quiz 4:** To eliminate x from the following simultaneous equations:

7x + 3y = 13 |
(1) |
---|---|

−2x + 5y = 8 |
(2) |

you can multiply (1) by −2 and (2) by 7. Which of the following equations for y will this procedure eventually yield?

(a)29y = 82Incorrect - please try again!

(b)29y = 30Correct - well done!

(c)41y = 82Incorrect - please try again!

(d)7y = −56Incorrect - please try again!

**Explanation:** Multiplication by −2 and 7 respectively yields:

−14x − 6y = −26 |
(3) |
---|---|

−14x + 35y = 56 |
(4) |

Subtracting (4) from (3) cancels the xs and yields:

−6y − 35y |
= | −26 − 56 |

−41y |
= | −82 |

41y |
= | 82 |

This implies that y = 2 and on substitution into (1) we obtain x = 1. These answers can then be checked by substituting into (2).

**Exercise 3:** Solve the following pairs of equations by first eliminating x:

(a)

3x + 4y = 10 |
(a1) |
---|---|

2x + 5y = 9 |
(a2) |

**Solution:** Multiply (a1) by 2 and (a2) by 3:

6x + 8y = 20 |
(a3) |
---|---|

6x + 15y = 27 |
(a4) |

The coefficient of x is now the same and subtracting (a3) from (a4) yields an equation in y alone:

15y − 8y |
= | 27 − 20 |

7y |
= | 7 |

so y = 1. Inserting this into (a1) yields 3x + 4 = 10, which implies that 3x = 6 and so x = 2. Check x = 2 , y = 1 by substitution into (a2)!

(b)

3x − 2y = 9 |
(b1) |
---|---|

-x + 3y = −3 |
(b2) |

**Solution:** Multiply (b1) by −1 and (b2) by 3:

-3x + 2y = −9 |
(b3) |
---|---|

-3x + 9y = −9 |
(b4) |

Subtracting (b4) from (b3) gives −7*y* = 0,
so that *y* = 0.

Inserting this into (b1) yields x = 3. The solution, x = 3, y = 0, should be checked by substitution into (b2):

−*x* + 3*y* = −3 + 0 ✓

(c)

2x − y = 5 |
(c1) |
---|---|

3x + 4y = 2 |
(c2) |

**Solution:** Multiply (c1) by 2 and (c2) by 3:

6x − 3y = 15 |
(c3) |
---|---|

6x + 8y = 4 |
(c4) |

and subtracting (c4) from (c3) gives −11*y* = 11, so *y* = −1.

Inserting this into the initial equation yields:

2x + 1 |
= | 5 |

2x |
= | 4 |

x |
= | 2 |

Now check that *x* = 2 , *y* = −1, by substitution into (c2)!

(d)

5x + 7t = 8 |
(d1) |
---|---|

7x - 4t = 25 |
(d2) |

**Solution:** Multiply (d1) by 7 and (d2) by 5:

35x + 49t = 56 |
(d3) |
---|---|

35x - 20t = 125 |
(d4) |

Subtracting (d4) from (d3) gives:

49t + 20t |
= | 56 − 125 |

69y |
= | -69 |

so t = −1. Inserting this into (d1) yields:

5x - 7 |
= | 8 |

5x |
= | 15 |

so we get *x* = 3 , *t* = −1. Check this by substitution into (d2)!

**Quiz 5:** Choose the solution of the following simultaneous equations:

^{1}⁄_{2}x + 2y = 3 |
(1) |
---|---|

2x + 3y = 7 |
(2) |

(a) *x* = ^{1}⁄_{2}, *y* = 2 Incorrect - please try again!

(b) *x* = −^{1}⁄_{2}, *y* = 2 Incorrect - please try again!

(c) *x* = 4, *y* = 0 Incorrect - please try again!

(d) *x* = 2, *y* = 1 Correct - well done!

**Explanation:** It is easiest here to multiply (1) by 4 and then subtract (2) from it. In this way we do not have unnecessary fractions. We find:

2x + 8y = 12 |
(3) |
---|---|

2x + 3y = 7 |
(4) |

Subtracting (4) from (3) cancels the xs and yields:

5y |
= | 5 |

y |
= | 1 |

Substituting this into (3) yields x = 1.
The solution, *x* = 2, *y* = 1, can be checked by substitution into (2).

Choose the solutions from the options given

1.If *x* + *y* = 1 and *x* − *y* = 3,
what are x and y?

(a)*x* = 2, *y* = −1

(b)*x* = −1, *y* = 2

(c)*x* = 2, *y* = 2

(d)*x* = 4, *y* = 1

2.To eliminate x from the following equations, *ax* + 2*y* = 4 and 3*x* − 2*ay* = −17, what should they be multiplied by?

(a)4 & −17

(b)*a* & 3

(c)3 & *a*

(d)2 & −2*a*

3.What is the solution of 3*x* + 2*y* = 1 and 2*x* + 3*y* = −1 ?

(a)*x* = 3, *y* = −4

(b)*x* = 5, *y* = 3

(c)*x* = −3, *y* = 5

(d)*x* = 1, *y* = −1

4.Given 2*x* − 3*y* = 1 and 3*x* − 2*y* = 4, what are x and y?

(a)*x* = 2, *y* = 1

(b)*x* = −1, *y* = −1

(c)*x* = 1, *y* = 2

(d)*x* = 3, *y* = −2

Example 2 illustrates the central idea of the method which is to combine the two equations so as to get a single equation for one variable and then use this to find the other unknown.