3. Equations and Inequalities

iGCSE (2021 Edition)

Lesson

In order to solve two equations with multiple variables simultaneously, we need a way to combine the information in both equations into a single equation with only one variable.

The substitution method achieves this by setting the values of the variables to be equal across both equations and then substituting the value of one variable into the other.

Another way to achieve this is using the elimination method, which takes a combination of the two equations such that all but one variable is eliminated.

Combining equations relies on the fact that the left and right-hand sides of any given equation are equal. Because they are equal, we can make expressions out of the two equations that still hold true.

For example: if we know that:

- Equation 1: $a=b$
`a`=`b` - Equation 2: $c=d$
`c`=`d`

Then we can combine these two equations to also know that:

- Equation 1 $+$+ Equation 2: $a+c=b+d$
`a`+`c`=`b`+`d` - Equation 1 $-$− Equation 2: $a-c=b-d$
`a`−`c`=`b`−`d`

We made these two new equations by adding or subtracting one equation from the other.

As well as simply adding or subtracting equations from each other, we can also take multiples of equations before combining them. As such, these equations must also hold true:

- $3$3$\times$×Equation 1 $-$− $4$4$\times$×Equation 2: $3a-4c=3b-4d$3
`a`−4`c`=3`b`−4`d` - $\frac{1}{2}$12$\times$×Equation 1 $+$+ $3$3$\times$×Equation 2: $\frac{1}{2}a+3c=\frac{1}{2}b+3d$12
`a`+3`c`=12`b`+3`d`

To solve equations simultaneously, we want to combine equations such that the new equation only has one variable.

Consider the equations:

- Equation 1: $3x+2y=6$3
`x`+2`y`=6 - Equation 2: $2x-2y=14$2
`x`−2`y`=14

In this case, we can see that the coefficients of the $y$`y`-terms in the equations are equal but with opposite signs. As such, adding the equations together would eliminate the variable $y$`y` like so:

$3x+2y+2x-2y$3x+2y+2x−2y |
$=$= | $6+14$6+14 |
Equation 1 $+$+ Equation 2 |

$3x+2x+2y-2y$3x+2x+2y−2y |
$=$= | $20$20 |
Collect constant terms |

$5x$5x |
$=$= | $20$20 |
Collect like terms |

$x$x |
$=$= | $4$4 |
Solve for $x$ |

As we saw in the second line of working out, the $y$`y`-terms cancelled each other out to give us an equation in terms of only $x$`x`. This allowed us to solve for the value of $x$`x`, which we can then substitute back into either equation to solve for the value of $y$`y`.

$3\times4+2y$3×4+2y |
$=$= | $6$6 |
Substitute $x=4$ |

$12+2y$12+2y |
$=$= | $6$6 |
Perform the multiplication |

$2y$2y |
$=$= | $-6$−6 |
Subtract $12$12 from both sides |

$y$y |
$=$= | $-3$−3 |
Divide both sides by $2$2 |

By adding the two equations together, we could eliminate the $y$`y`-terms, allowing us to solve for $x$`x`, which we substituted back in to solve for $y$`y`.

However, not every pair of equations will have coefficients that nicely match up for easy elimination of variables. In some cases, we will need to also multiply the equations before combining them.

Solve the equations simultaneously using the elimination method:

- Equation 1: $5x-2y=7$5
`x`−2`y`=7 - Equation 2: $2x-3y=-6$2
`x`−3`y`=−6

**Think:** We want eliminate either the $x$`x` or $y$`y`-terms when combining the equations. Since neither has matching coefficients, we will need to multiply the equations before combining them.

Suppose we want to eliminate the $y$`y`-terms. We can make the coefficients match if we multiply Equation 1 by $3$3 and Equation 2 by $2$2.

**Do:** Multiplying the equations by the required values gives us:

- $3$3$\times$×Equation 1: $15x-6y=21$15
`x`−6`y`=21 - $2$2$\times$×Equation 2: $4x-6y=-12$4
`x`−6`y`=−12

Now the $y$`y`-terms have equal coefficients with the same sign. As such, we can eliminate them by subtracting one equation from the other.

$15x-6y-\left(4x-6y\right)$15x−6y−(4x−6y) |
$=$= | $21-\left(-12\right)$21−(−12) |
$3$3$\times$×Equation 1$-$− $2$2$\times$×Equation 2 |

$15x-6y-4x+6y$15x−6y−4x+6y |
$=$= | $21+12$21+12 |
Remove the brackets |

$11x$11x |
$=$= | $33$33 |
Collect like terms |

$x$x |
$=$= | $3$3 |
Divide both sides by $11$11 |

Now that we know the value for $x$`x`, we can substitute it back into one of our starting equations and solve for $y$`y`.

$5\times3-2y$5×3−2y |
$=$= | $7$7 |
Substitute $x=3$ |

$15-2y$15−2y |
$=$= | $7$7 |
Perform the multiplication |

$-2y$−2y |
$=$= | $-8$−8 |
Subtract $15$15 from both sides |

$y$y |
$=$= | $4$4 |
Divide both sides by $-2$−2 |

Therefore, the solution to Equations 1 and 2 simultaneously is:

$x=3,y=4$`x`=3,`y`=4

Given the following equations, we want to solve for $x$`x` and $y$`y` using the elimination method.

Equation 1 | $7x+9y=-41$7x+9y=−41 |

Equation 2 | $5x-18y=6$5x−18y=6 |

Notice that Equation 1 has a $9y$9

`y`term and Equation 2 has a $-18y$−18`y`term. How can we combine the equations to eliminate the $y$`y`-terms?Equation 1 $-$−$2$2$\times$× Equation 2

A$2$2$\times$× Equation 1 $-$− Equation 2

B$2$2$\times$× Equation 1 $+$+ Equation 2

CEquation 1 $+$+ Equation 2

DEquation 1 $-$−$2$2$\times$× Equation 2

A$2$2$\times$× Equation 1 $-$− Equation 2

B$2$2$\times$× Equation 1 $+$+ Equation 2

CEquation 1 $+$+ Equation 2

DSolve for $x$

`x`by adding Equation 2 to $2$2 times Equation 1.Enter each line of working as an equation.

Substitute $x=-4$

`x`=−4 into either of the equations and solve for $y$`y`.Enter each line of working as an equation.

Given the following equations, we want to solve for $x$`x` and $y$`y` using the elimination method.

Equation 1 | $3x+11y=-48$3x+11y=−48 |

Equation 2 | $-2x+7y=-\frac{76}{3}$−2x+7y=−763 |

Notice that Equation 1 has a $3x$3

`x`term and Equation 2 has a $-2x$−2`x`term. How can we combine the equations to eliminate the $x$`x`-terms?$2$2$\times$× Equation 1 $+$+ $3$3$\times$×Equation 2

AEquation 1 $+$+ $3$3$\times$× Equation 2

B$2$2$\times$× Equation 1 $-$− Equation 2

C$2$2$\times$× Equation 1 $-$− $3$3$\times$×Equation 2

D$2$2$\times$× Equation 1 $+$+ $3$3$\times$×Equation 2

AEquation 1 $+$+ $3$3$\times$× Equation 2

B$2$2$\times$× Equation 1 $-$− Equation 2

C$2$2$\times$× Equation 1 $-$− $3$3$\times$×Equation 2

DSolve for $y$

`y`by adding $2$2 times Equation 1 to $3$3 times Equation 2.Enter each line of working as an equation.

Substitute $y=-4$

`y`=−4 into either of the equations and solve for $x$`x`.Enter each line of working as an equation.

For two numbers, $x$`x` and $y$`y`:

- two times the first number is added to the second number to get $27$27, and
- the difference between five times the first number and the second number is $43$43.

We want to find the values of these two numbers.

Set up two equations by letting $x$

`x`and $y$`y`be the two numbers.Use $x$

`x`as the first of the two numbers.Sum equation: $\editable{}$ Difference equation: $\editable{}$ First solve for $x$

`x`by eliminating the $y$`y`-terms.Equation 1 $2x+y=27$2 `x`+`y`=27Equation 2 $5x-y=43$5 `x`−`y`=43Now solve for $y$

`y`.

Solve simple simultaneous equations in two unknowns by elimination or substitution.