Question

Solve the following system of linear equations by graphing:4x + 4y = 206x + 6y = 6

Answer 1

Given the system of equations, we shall start by expresing them in slope-intercept form as follows;

`\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} \\ (4y)/(4)=(20-4x)/(4) \\ y=(20)/(4)-(4x)/(4) \\ y=5-x \\ y=-x+5---(1) \end{gathered}`

For the second equation, we would have;

`\begin{gathered} 6x+6y=6 \\ S\text{ubtract 6x from both sides; } \\ 6y=6-6x \\ \text{Divide both sides by 6} \\ (6y)/(6)=(6-6x)/(6) \\ y=(6)/(6)-(6x)/(6) \\ y=1-x \\ y=-x+1---(2) \end{gathered}`

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).