# Thread: Word Problem - Simultaneous Equation?

1. ## Word Problem - Simultaneous Equation?

Hi guys, I'm struggling to set up this problem into numbers

If a teacher can place her students four to a bench, there will be three students on the final bench. But, if five children are placed on each bench, there will be four students on the last bench.

What is the smallest number of the children the class could have?
I'm thinking x=number of benches and y=different number of benches, so

4x+3=5y+4

Not really sure where to go from there? Any light shed would be greatly appreciated!

2. With 5 children on a bench + 4 left over, the number of students must have a 4 or 9 as the last digit. But with 4 children on a bench + 3 left over, it's impossible for the last digit to be 4, so it must be 9. The smallest number that ends in a 9 that fulfills the conditions is 19.

01

3. Hello, BoulderBrow!

Another approach . . .

If a teacher places her students 4 to a bench, there will be 3 students on the final bench.
But if 5 children are placed on each bench, there will be 4 students on the last bench.

What is the smallest number of the children the class could have?

I'm thinking $\displaystyle x$ = number of benches and $\displaystyle y$ = different number of benches.

So: .$\displaystyle 4x+3 \:=\:5y+4$ . . . . Good!
Solve for $\displaystyle y\!:\;\;y \:=\:\frac{4x-1}{5}$

Since $\displaystyle y$ is a positive integer, $\displaystyle 4x-1$ must be divisible by 5.

Trying $\displaystyle x \:=\:1,2,3,\hdots$ we find the first occurence is: $\displaystyle x = 4.$
. . And hence: $\displaystyle y = 3$

Therefore, the number of students is: .$\displaystyle 4(4)+3 \:=\:5(3)+4 \:=\:\boxed{19}$