3. Equations and Inequalities

iGCSE (2021 Edition)

Lesson

In order to solve for a variable in an equation, we need to be able to isolate that variable. If we are given a single equation with two unknown variables, it is impossible to solve for either value.

However, when we are given two equations, both with the same two variables, we are able to solve for both variables.

When solving equations simultaneously, we are looking for some values for our variables which make both equations true. This means that in two equations in terms of $x$`x` and $y$`y`, each variable can be considered to be some fixed value for both equations- even if we don't know what that value is yet!

By assuming that the values for $x$`x` and $y$`y` in one equation are the same as for the other equation, we are able to substitute the value of a variable from one equation into the other.

Consider the two equations:

- Equation 1: $y=5x+11$
`y`=5`x`+11 - Equation 2: $y=3x+5$
`y`=3`x`+5

If we solve these equations simultaneously, the $x$`x` and $y$`y`-values of the solution will be the same for both equations.

As such, we can assume that the variable $y$`y` in Equation 1 will have the same value as the variable $y$`y` in Equation 2.

From Equation 1, we know that $y$`y` is equal to $5x+11$5`x`+11. Therefore, the variable $y$`y` in Equation 2 will also be equal to $5x+11$5`x`+11. As such, we can rewrite Equation 2 to be:

$5x+11=3x+5$5`x`+11=3`x`+5

We have substituted the value of $y$`y` from Equation 1 into Equation 2.

Now that we have an equation with only one variable, we can solve it to find $x$`x`. This will give us:

$x=-3$`x`=−3

Since we are assuming that the $y$`y`-values are equal for both equations, we are also assuming that the $x$`x`-values are the same for both equations.

This means that $x=-3$`x`=−3 is the $x$`x` -value of the solution to both equations simultaneously.

We can find the corresponding $y$`y`-value by substituting $x=-3$`x`=−3 into either of the equations. Suppose we substitute it into Equation 1, which will give us:

$y=5\times\left(-3\right)+11$`y`=5×(−3)+11

Solving this tells us that the $y$`y`-value for the solution is:

$y=-4$`y`=−4

To double check that $x=-3$`x`=−3, $y=-4$`y`=−4 is indeed the solution to Equations 1 and 2 simultaneously we can substitute the values into both equations.

- Equation 1: $-4=5\times\left(-3\right)+11$−4=5×(−3)+11
- Equation 2: $-4=3\times\left(-3\right)+5$−4=3×(−3)+5

Since both equations hold true, we have confirmed that $x=-3$`x`=−3, $y=-4$`y`=−4 is the simultaneous solution to the equations.

In the example above, both equations already had $y$`y` as the subject so it was easy to substitute it from one equation into the other. When this is not the case, we can choose one of the variables to isolate in one equation so that we can substitute it into the other.

We want to solve the following system of equations using the substitution method.

- Equation 1: $5x+2y=22$5
`x`+2`y`=22 - Equation 2: $2x+5y=-8$2
`x`+5`y`=−8

**Think:** To solve the equations simultaneously using the substitution method, we want to isolate one of the variables in one of the equations, then substitute it into the other.

**Do:** We can start by isolating $x$`x` in Equation 1. By subtracting $2y$2`y` from both sides and then dividing both sides of the equation by $5$5, we get:

$x=\frac{22}{5}-\frac{2y}{5}$`x`=225−2`y`5

Substituting this into Equation 2 gives us:

$2\left(\frac{22}{5}-\frac{2y}{5}\right)+5y=-8$2(225−2`y`5)+5`y`=−8

After expanding the brackets and collecting the like terms on either side, we get:

$\frac{21y}{5}=\frac{-84}{5}$21`y`5=−845

As such, we get:

$y=-4$`y`=−4

Substituting this into Equation 2 gives us:

$2x+5\times\left(-4\right)=-8$2`x`+5×(−4)=−8

So we get:

$x=6$`x`=6

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

$x=6,y=-4$`x`=6,`y`=−4

We want to solve the following system of equations using the substitution method.

Equation 1 | $y=4x-28$y=4x−28 |

Equation 2 | $3x+5y=44$3x+5y=44 |

First solve for $x$

`x`.Now, solve for $y$

`y`.

We want to solve the following system of equations using the substitution method.

Equation 1 | $-7p-2q=-\frac{99}{40}$−7p−2q=−9940 |

Equation 2 | $-21p-10q=-\frac{85}{8}$−21p−10q=−858 |

First solve for $q$

`q`.Now solve for $p$

`p`.

A man is four times as old as his son. Six years ago the man was ten times as old as his son.

We want to find their present ages.

Let $x$

`x`and $y$`y`be the ages of the man and his son respectively.Use the fact that the man is four times as old as his son to set up Equation 1 in the form $x=\editable{}$

`x`=.Use the fact that four years ago the man was ten times as old as his son to set up Equation 2.

Write the equation in the form $ax+by=c$

`a``x`+`b``y`=`c`, where $a$`a`is positive.First solve for $y$

`y`to find the age of the son.Equation 1 $x=4y$ `x`=4`y`Equation 2 $x-10y=-54$ `x`−10`y`=−54Now solve for $x$

`x`to find the age of the man.Equation 1 $x=4y$ `x`=4`y`Equation 2 $x-10y=-54$ `x`−10`y`=−54

Use substitution to form and solve a quadratic equation in order to solve a related equation.

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