# Fun with simultaneous equations

**Example 1**

**What is a sheep worth?**

The way to tackle this question is to substitute the sheep for a letter. So the equation above becomes:

`3(5x+1)=18`

Multiply the `3` into the brackets:

`3times5x+3times1=18`

`15x+3=18`

Subtract `3` from both sides to help get `x` on its own:

`15x+cancel(3)-cancel(3)=18-3`

`15x=15`

Divide both sides by `15` to get `x` on its own:

`(cancel(15)x)/cancel(15)=15/15`

`x=15/15`

`x=1`

**Answer:** The sheep is worth 1.

**Example 2**

**What is a sheep worth and what is a cow worth?**

The way to tackle this question is to substitute both the sheep and the cow with letters. So the equations become:

`s+c=3` ...equation 1

`s-c=1` ...equation 2

Take the first equation and subtract `s` from both sides to get `c` on its own:

`s-s+c=3-s`

`c=3-s`

Now substitute `c` into equation 2:

`s-(3-s)=1`

There is a 1 behind the minus sign but mathematicians usually hide it, but in this case we will put it back:

`s-1(3-s)=1`

Multiply the `-1` into the brackets:

`s-3+s=1`

Add the two `s`'s:

`-3+2s=1`

Add `3` to both sides to help get `s` on its own:

`-cancel(3)+cancel(3)+2s=1+3`

`2s=4`

Divide both sides by `2` to get `s` on its own:

`(cancel(2)s)/cancel(2)=4/2`

`s=4/2`

`s=2`

Now we know that `s=2` we can plug that back into equation 1 or 2 to get the value of `c`.

Try equation 1:

`s+c=3`

`2+c=3`

Take 2 from both sides to get `c` on its own:

`cancel(2)-cancel(2)+c=3-2`

`c=1`

**Answer:** The sheep is worth 2 and the cow is worth 1.

**Example 3**

**What is the value of the question mark?**

The way we tackle this question is to substitute each symbol for a letter i.e. sign = `s`, house = `h`, key = `k`.

`h+h+h=15` ...equation 1

`h+s+s=13` ...equation 2

`s+k+k=8` ...equation 3

`s+ktimesh=?` ...equation 4

Immediately we can see that that equation 1 is the easiest to tackle:

`h+h+h=15`

Add the `h`'s together and we get:

`3h=15`

Divide both sides by `3` to get `h` on its own:

`(cancel(3)h)/cancel(3)=15/3`

`h=15/3`

`h=5`

Substitute `h=5` into equation 2 and we get:

`h+s+s=13`

`5+s+s=13`

`5+2s=13`

Minus `5` from both sides to help get `s` on its own:

`cancel(5)-cancel(5)+2s=13-5`

`2s=8`

Divide both sides by `2` to get s on its own:

`(cancel(2)s)/cancel(2)=8/2`

`s=8/2`

`s=4`

Substitute `s=4` into equation 3 and we get:

`s+k+k=8`

`4+k+k=8`

`4+2k=8`

Minus `4` from both sides to help get `k` on its own:

`cancel(4)-cancel(4)+2k=8-4`

`2k=4`

Divide both sides by `2` to get `k` on its own:

`(cancel(2)k)/cancel(2)=4/2`

`k=4/2`

`k=2`

Substitute values of `h=5`, `s=4` and `k=2` into equation 4 and we get:

`s+ktimesh=?`

`4+2times5=?`

`4+10=?`

`?=14`

**Answer:** The value of the question mark is `14`.

**Example 4**

**Find the value of a triangle and a circle.**

The way to tackle this problem is to substitute each symbols for a letter i.e. triangle`=x` and circle`= y`.

`x+x+y=7` ...equation 1

`x+y+y=5` ...equation 2

`x+x+y+y+y+y=10` ...equation 3

Which then becomes:

`2x+y=7` ... equation 1

`x+2y=5` ...equation 2

`2x+4y=10` ...equation 3

We can see that equations 1 and 3 both have a `2x` in them.

Tackling equation 1:

`2x+y=7`

Take `y` from either side to get `2x` on its own:

`2x+cancel(y)-cancel(y)=7-y`

`2x=7-y`

Substitute this into equation 3:

`2x+4y=10`

`7-y+4y=10`

`7+3y=10`

Take `7` from both sides to help get `y` on its own:

`cancel(7)-cancel(7)+3y=10-7`

`3y=3`

Divide both sides by `3` to get `y` on its own:

`(cancel(3)y)/cancel(3)=3/3`

`y=3/3`

`y=1`

Substitute `y=1` into equation 1 and we get:

`2x+y=7`

`2x+1=7`

Take `1` from both sides to help get `x` on its own:

`2x+cancel(1)-cancel(1)=7-1`

`2x=6`

Divide both sides by `2` to get `x` on its own:

`(cancel(2)x)/cancel(2)=6/2`

**Answer:**

So `y=1` and `x=3`.

or circle`=1` and triangle `=3`.

Note: We didn't even need to use equation 2.