Question

Given that x and y are positive integers and (x·2)+x(2^2)+(y·2^3)+(y·2^4)=42, what is the value of xy?

Answer 1

I am gonna try my best to explain this to you :)

Firstly what you need to do is simplify the equation

$(x*2)+x(2^2)+(y*2^3)+(y*2^4)=42\\2x+4x+8y+16y=42\\6x+24y=42$

Now it looks a little less complicated the next step is to isolate y, in order to do that we solve the equation for y

$6x+24y=42$

$24y=42-6x\\y=\frac{42-6x}{24}$ Simplify it $\frac{7-x}{4}$

Now substitute y for $\frac{7-x}{4}$

$6x+\frac{7-x}{4}=42\\4(6x)+7-x=42*4\\24x+7-x=168\\23x+7=168\\23x=168-7\\23x=161\\x=\frac{161}{23}\\x=7$

Now substitute x for 7 and solve for y

$6(7)+24y=42\\42+24y=42\\24y=42-42\\24y=0\\y=\frac{0}{42}\\y=0$

x=7 and y=0

Hope this helps :)