![3x+2y=12](https://img.qammunity.org/2023/formulas/mathematics/college/6ck0r6kabeudhn3urup5z3g2tbzj7h6xkt.png)
a. x- intercept: the point at which the graph crosses the x-axis, when the y-value is 0.
To find the x- intercept in the given equation substitute the y by 0 and solve for x:
![\begin{gathered} 3x+2(0)=12 \\ 3x+0=12 \\ 3x=12 \\ \\ \text{Divide both sides of the equation into 3:} \\ (3)/(3)x=(12)/(3) \\ \\ x=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oqftma688nd7zumph2gwdoj6a1sgk6s3lk.png)
Then, the x-intercept is ( 4, 0 )
b. y-intercept: the point at which the graph crosses the y-axis, when the x-value is 0.
To find the y-intercept in the givne equation substitute the x by 0 and solve for x:
![\begin{gathered} 3(0)+2y=12 \\ 0+2y=12 \\ 2y=12 \\ \\ \text{Divide both sides of the equation into 2:} \\ (2)/(2)y=(12)/(2) \\ \\ y=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q0mzbgzu65hl7o29qmuy9sz6navlix7y1j.png)
Then, the y-intercept is ( 0 , 6 )
c. Graph: Use the x-intercept and y-intercept, put it in the plane and then draw a line that passes trough those points: