So,
For number one, there will be an infinite number of solutions, as there are two placeholders.
3x - 2y = 6
It seems we can use the intercept method.
3x - 2(0) = 6
3x = 6
Divide both sides by 3
x = 2
This is one solution: (2,0)
3(0) - 2y = 6
-2y = 6
Divide both sidees by -2
y = -3
This is another solution: (0,-3)
You should now be able to draw the line.
If you want intercept form for this equation, just manipulate the equation in order to get the y = mx + b form.
Subtract 3x from both sides
-2y = -3x + 6
Divide both sides by -2
\(y = \frac{3}{2} x + 6\)
For the second problem, simply manipulate the equation.
\(\frac{1}{3} (6x - 9) = \frac{3}{4} (4x - 8)\)
Distribute
\(2x - 3 = 3x - 6\)
Subtract 2x from both sides
\(-3 = x - 6\)
Add 6 to both sides
\(3 = x\)
For the third problem, do the same thing.
\(\frac{3(2x-1)}{4} = x + \frac{7}{2}\)
We must first multiply both sides by 4.
\(3(2x-1) = 4x + 14\)
Distribute
\(6x - 3 = 4x + 14\)
Subtract 4x from both sides
\(2x - 3 = 14\)
Add 3 to both sides
\(2x = 17\)
Divide both sides by 2
\(x = \frac{17}{2} \ or\ 8 \frac{1}{2}\)
