Real Life System of Equations!

l + e + p = 120.

l = 3e

p = 6e

So, e + 3e + 6e = 120 So e = 12.

So (l ÷ p) + e (6e ÷ 3e) + 12 2 + 12 = 14

Translating the given information into equations

• $l+e+p=120$
• $l=3e$
• $e=\frac{p}{6}$

Manipulating them to find expressions of $e$ that we can substitute in the first equation above

• $e=e$
• $l=3e$
• $6e=\frac{p}{6}*6=\frac{6p}{6}=\frac{p}{1}=p$

Substituting in the first equation, then finding the value of $e$

• $p+l+e=6e+3e+e=9e+e=10e=120$
• $\frac{10e}{10}=\frac{e}{1}=e=\frac{120}{10}=12$

Substituting values of e to find $l$ and $p$

• $e=12$
• $l=3e=3*12=36$
• $p=6e=6*12=72$

Checking the values for $e$ , $l$ and $p$

• $36+12+72=120$

Plugging these values into the final expression

• $\frac{p}{l}+e=\frac{72}{36}+12=2+12=\boxed{14}$

Let $l =$ Linda , $E =$ Ella & $p =$ Peter

We are told that

$l + E + p = 120$

And that

$l = 3E$

$E = \frac {p}{6}$

Substituting these values in gives

$3E + \frac {p}{6} + p = 120$

Substituting the second expression in again gives

$3(\frac {p}{6}) + \frac {p}{6} + p = 120$

Let's simplify

$\frac {5p}{3} = 120$

And now solve

$p = 72$

Since we already have a equation for $E$ in terms of $p$ , we just need to find an equation for $l$ in terms of $p$

$l = \frac {p}{2}$

Done, now lets substitute them into the final equation

$\frac {(72)}{\frac {(72)}{2}} + \frac {(72)}{6}$

Now we simplify

$\frac {72}{36} + 12 = 2 + 12 = 14$

So the equation $\frac {p}{l} + E$ equals $14$