For the word problem below, there are 4 variables to solve for, but I can only identify 3 equations given the data. Through trial and error, you can easily figure out the answer, but I am wondering if there is an implied equation I'm not seeing. In other words, how can this problem be solved using algebra rather than a bit of trial and error?
Problem
Ian has been saving money for his summer trip to his grandparents' house. He has saved 18 bills of varying denominations, including \$20s, \$10s, \$5s, \$1s. He's saved a total of \$126. He has the same number of \$10s as he has \$1s. How many of each bill does Ian have?
Equations
w + x + y + z = 18
w + 5x + 10y + 20z = 126
w = y
[legend // w: number $1s, x: number of $5s, y: number of $10s, z: number of $20s]
Answer
$1s: 6
$5s: 4
$10s: 6
$20s: 2
$$w+x+y+z=18;\;w+5 x+10 y+20 z=126;\;y=w$$ then $$11w+5x+20z=126;\;2w+x+z=18$$ multiply the second equation by $-5$ and add to the first $$-10w-5x-5z=-90;\;w+15z=36$$ so $$z=\frac{36-w}{15}$$
As the numbers are positive integers, $36-w$ must be a multiple of $15$. Furthermore all unknowns are less than $18$
$36 - w=15$ leads to $w=21$ which is greater than $18$
$36-w=30$ leads to $w=6;\;y=6;\;x=4;\;z=2$
$36-w=45$ leads to a negative w.
Thus the solution found above is unique.