# Matchless

There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?

There is a particular value of $x$, and a value of $y$ to go with it, which make all five expressions equal in value.

Can you find that $x$, $y$ pair?

Image

Did you have more information than you needed to solve the problem?

For a start: $2x + 3y - 20$ equals $ 5x - 2y +38$ ...

Well done to all the people who combined the expressions in such a way as to eliminate the $x$ or $y$ variable. For example Sakshi from Singapore did this :

$(2x + 3y - 20) + (4x + 5y - 72) = 6x + 8y - 92 $

and

$(5x - 2y + 38) + (x - 4y + 108) = 6x - 6y + 146 $

These two results must both equal the same value.

And since $6x$ is in both expressions, contributing the same to each expression whatever its actual value, $8y - 92$ must match $146 - 6y $

from which it follows that $14y = 238$ with $y = 17 $

x can then be found with a similar approach :

(2x + 3y - 20) + (5x - 2y + 38) = 7x + y + 18

(4x + 5y - 72) + (x - 4y + 108) = 5x + y + 36

And since 7x + y + 18 matches the value of 5x + y + 36 , x must be 9

Jennifer from Our Lady's Convent had another good approach :

Set any two of the five expressions equal to each other.

$2x + 3y - 20 = 5x - 2y + 38 $

and then simplify the expression to get :

$-3x + 5y - 58 = 0 $

Set any other pair of expressions equal to each other

$4x + 5y - 72 = 3x - y + 39 $

then again simplifyingto get :

$x + 6y - 111 = 0 $

I now have two simultaneous equations, and as there are two unknown values in each, I will make $x$ the subject of one of them, and substitute it into the other.

Substitute $x = 111 - 6y$ into $-3x + 5y - 58 = 0$

$-3(111 - 6y) + 5y - 58 = 0 $

$-333 + 18y + 5y - 58 = 0 $

$23y = 391$ and so $y = 17$

Substitute $y = 17$ into $-3x + 5y - 58 = 0$

to get : $-3x + (5 x 17) - 58 = 0 $

so $x = 9$

Jennifer then checked the whole thing by substituting $y = 17$ and $x = 9$ into all expressions to find that for that combination of values each expression produced $49$ . Brilliant!

Alice from Colyton Grammar School explains that in fact only $3$ of the expressions are needed, any three will do. Here's her method :

To find the values of $x$ and $y$ for all the expressions to be equal let: $2x + 3y - 20 = 5x - 2y + 38$

simplify to get $3x - 5y = -58$ ......($1$)

similarly let: $4x + 5y - 72 = 5x - 2y + 38 $

simplify to get $x - 7y = -110$ ......($2$)

solve ($1$) and ($2$) simultaneously

$(1) - 3*(2)$ leads to $16y = 272$ so that $y = 17$

Sub $y = 17$ into ($2$)

$x = -110 + 7y = 9 $

So only $3$ equations are needed to solve the problem as only two equations are needed to solve an equation with two variables however, the additional equations given can be used to check if the $x$ and $y$ values are correct.

A teacher's account of working with students on this problem has been added to the Notes section, do have a look.

*This printable worksheet may be useful: Matchless.*

Listen to Jenny and Graeme talking together about the problem. [ audio ]

Encounters with simultaneous equations can become over-familiar, routine experiences for students. This type of problem causes a "stop and think" moment, requires some problem-solving ingenuity, and leads into a consideration of redundancy of information. Students might make a start by substituting some arbitrary $x$, $y$ values to get a feel for the problem and to grasp that the five expressions don't generally take the same value.

This is in contrast to expressions that are identities, for example $2 (x+y) - 3(x-y )$ and $5y - x$, where the two expressions take the same value for any $x$, $y$ combination. This idea is worth some discussion.

Questions or prompts:

#### For a start:

#### $2x + 3y - 20$ equals $ 5x - 2y +38$ ...

#### Could you find an $x$, $y$ pair that works for two, for three, or for four of the expressions but not for all of them ?

#### Further ideas:

Make up a similar problem of your own.Or extending that : can you create a similar problem with an "odd one out"? That is, one expression which does not equal the other four, which are equal for some specific $x$, $y$ pair.

The following interesting account was sent in by a class teacher working withYear 8s in Maths Club at St Albans High School for Girls

Becky worked as follows:

$2x + 3y -20 = 4x + 5y -72$

($-2x$ to each side)

$3y -20 = 2x + 5y - 72$

($+72$ to each side)

$3y + 52 = 2x + 5y$

($-3y$ to each side)

$52 = 2x + 2y$

Then, $5x - 2y + 38 = x - 4y + 108$

($-x$ from each side)

$4x - 2y + 38 = -4y + 108$

($+4y$ to each side)

$4x + 2y + 38 = 108$

( $-38$ to each side)

$4x + 2y = 70$

Becky looked at the difference between these two equations and deduced

$2x = 18, so x = 9$

Then using one of her equations, and substituting $x = 9$ she found $y = 17$.

Ele and Sarah reached the same conclusion but started out by looking at

all the possible pairs of expressions. Then they selected the following two

as easiest to work with:

From $2x + 3y - 20 = 4x + 5y - 72$

They deduced $26 = x + y$

From $2x + 3y - 20 = x - 4y + 108$

They deduced $128 = x + 7y$

From these two equations they deduced $6y = 128 - 26$

$6y = 102$

$y= 17$

Then from $x + y = 26$ they found $x = 9$

All three girls were then challenged to decide how much of the information

they needed to use to solve the problem. Their conclusion was only $3$

statements were needed; they could have used $A = B$ and $A = C$ to deduce the answer where $A,B,C$ label different expressions given.