Given the system of equations, we shall start by expresing them in slope-intercept form as follows;
[tex]\begin{gathered} The\text{ equation in slope-intercept form is;} \\ y=mx+b \\ 4x+4y=20\text{ would now be;} \\ \text{Subtract 4x from both sides;} \\ 4y=20-4x \\ \text{Divide both sides by 4} \\ \frac{4y}{4}=\frac{20-4x}{4} \\ y=\frac{20}{4}-\frac{4x}{4} \\ y=5-x \\ y=-x+5---(1) \end{gathered}[/tex]
For the second equation, we would have;
[tex]\begin{gathered} 6x+6y=6 \\ S\text{ubtract 6x from both sides; } \\ 6y=6-6x \\ \text{Divide both sides by 6} \\ \frac{6y}{6}=\frac{6-6x}{6} \\ y=\frac{6}{6}-\frac{6x}{6} \\ y=1-x \\ y=-x+1---(2) \end{gathered}[/tex]
With the two equations now written in slope-intercept form, the graph shall be as follows;
The red line is the graph of y = -x + 5, while the blue line is the graph of y = -x + 1
A we can see, both graphs have the same slope, that is, -1, and that means the lines are parallel. Hence, no solution exists for this system of equations (since the lines cannot intersect).
