# Thread: By using algebraic method, find the value of the smallest value

1. ## By using algebraic method, find the value of the smallest value

Question:
The subtraction of two positive numbers is $4$. When the two numbers are multiplied together, the resulting value is $21$. By using algebraic method, find the value of the smallest number

My workings:
let $x$ be the smallest number
let $x+4$ be the bigger number
$(x+4)-x=4$
$(x+4)$ x $x=21$
$x+4=4+x$
$(x^2+4x)/4 = 21/4$
$4(x^2+4x)=84$
$4x^2+16x=84$
$4x^2+16x-84=0$
$(x+7)$ or $(4x-12)$
$x+7=0$ or $4x-12=0$
$x=-7$ (N.A.) or $4x-12=0$
Therefore, $x=3$

Am I right?

Thanks

2. ## Re: By using algebraic method, find the value of the smallest value

Originally Posted by FailInMaths
Question:
The subtraction of two positive numbers is $4$. When the two numbers are multiplied together, the resulting value is $21$. By using algebraic method, find the value of the smallest number

My workings:
let $x$ be the smallest number
let $x+4$ be the bigger number
$(x+4)-x=4$
$(x+4)$ x $x=21$
$x+4=4+x$
$(x^2+4x)/4 = 21/4$
$4(x^2+4x)=84$
$4x^2+16x=84$
$4x^2+16x-84=0$
$(x+7)$ or $(4x-12)$
$x+7=0$ or $4x-12=0$
$x=-7$ (N.A.) or $4x-12=0$
Therefore, $x=3$

Am I right?

Thanks
Plug your numbers into the original equations and see if it works.
Given that your smaller number is 3 then your larger number is 7.

$7 - 3 = 4 \text{ and } 7 \times 3 = 21$

Since that is true and matches the question statement your answer is correct.

Your working confused me though, why didn't you just go from saying $x(x+4) = 21$ in the second line to $x^2+4x = 21$ and then $x^2+4x-21=0$ and solve that quadratic?

3. ## Re: By using algebraic method, find the value of the smallest value

Originally Posted by e^(i*pi)
Plug your numbers into the original equations and see if it works.
Given that your smaller number is 3 then your larger number is 7.

$7 - 3 = 4 \text{ and } 7 \times 3 = 21$

Since that is true and matches the question statement your answer is correct.

Your working confused me though, why didn't you just go from saying $x(x+4) = 21$ in the second line to $x^2+4x = 21$ and then $x^2+4x-21=0$ and solve that quadratic?
Oh, now that you say it. To be honest, I had been thinking for a few hours, so I am not really sure of what I am writing, thus want to have it checked here

Thanks for spotting the mistake, I will amend it.

Thanks