Question 271380
Set up:
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Let A = # of 1 bedroom apartments
Let B = # of 2 bedroom apartments
Let C = # of 3 bedroom apartments

Equation 1: {{{A + B + C = 438}}}
Equation 2: {{{A + 48 = C}}}
Equation 3: {{{4A - 24 = B}}}

Solution:
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Notice that all the variables are in terms of A.
We can substitute the values in terms of A into equation 1
Equation 1: {{{A + B + C = 438}}}
{{{A + (4A - 24) + (A + 48) = 438}}} Rewrite the equation
{{{A + 4A - 24 + A + 48 = 438}}} Combine like terms
{{{6A + 24 = 438}}} Subtract 24 from both sides
{{{6A = 414}}} Divide both sides by 6
{{{highlight(A = 69)}}}

Now plug 69 into equations 2 & 3 for A
Equation 2: {{{A + 48 = C}}}
{{{69 + 48 = C}}}
{{{highlight(117 = C)}}}
Equation 3: {{{4A - 24 = B}}}
{{{4*(69) - 24 = B}}}
{{{highlight(252 = B)}}}

Check you answers:
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Plug your values back into equation 1

Equation 1: {{{A + B + C = 438}}}
{{{69 + 252 + 117 = 438}}}
{{{highlight(438 = 438)}}}