6)
the equation is:
![3x-2y=-4](https://img.qammunity.org/2023/formulas/mathematics/college/x238ojq4bp1mhac2bb984u5n7kvhl8ri4b.png)
So we need two poinds to graph the line so I can replace X=0 so:
![\begin{gathered} -2y=-4 \\ y=(4)/(2) \\ y=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cxvbrxanetavv655fqpj9t3dqw6mys431j.png)
So the first coordinate is (0,2) and now we replace y=0 so:
![\begin{gathered} 3x=-4 \\ x=-(4)/(3) \\ x=1.33 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1n96r8nnyr5tljh1y3tyfj5rlbkcukrr67.png)
So the secon coordinate is (1.33,0), So now we can graph it:
7)
The equation is:
![y=\lvert x-5\rvert-3](https://img.qammunity.org/2023/formulas/mathematics/college/xiqaqy1i3isjhfu4vq5h587mghywk1p7k8.png)
This is an absolut value that is translate 5 units to the right and 3 units down, what menas the vertex is in the coordinate (5,-3), now we evaluate one point at the right and one at the left of the vertex so we evaluate it in x=0 and x=6
for x=0 will be:
![\begin{gathered} \lvert-5\rvert-3=y \\ 5-3=y \\ 2=y \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h5cszn1rhvnzux9dbsem8ywdv2hfohnujb.png)
Now for x=6
![\begin{gathered} \lvert6-5\rvert-3=y \\ 1-3=y \\ -2=y \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/odfwnz76cdyfnfgicm1pj7rqntlhlipsp6.png)
So the graph will be: