3x + 4y = 72
To find the y intercept, let x = 0 and we have
3(0) + 4y = 72
4y = 72 divide by 4
y = 18
To find the x intercept, let y =0 and we have
3x + 4(0) = 72
3x = 72 divide by 3
x = 24
3x+4y=72
I need the x and y intercepts.
\(\begin{array}{|rcll|} \hline 3x+4y &=& 72 \quad & | \quad :72 \\\\ \dfrac{3x+4y}{72} &=& 1 \\\\ \dfrac{3x}{72} + \dfrac{4y}{72} &=& 1 \\\\ \dfrac{x}{\dfrac{72}{3}} + \dfrac{y}{\dfrac{72}{4}} &=& 1 \\\\ \dfrac{x}{24} + \dfrac{y}{18} &=& 1 \\\\ \text{x-intercept}~ = 24, \text{if }~y=0\\ \text{y-intercept}~ = 18, \text{if }~x=0\\ \hline \end{array}\)