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'Matchless' printed from http://nrich.maths.org/
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 $
$(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
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 $49$ .
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.