1.solve the following equation for y 3x- 2y= 6
2. solve the following equation 1/3(6x-9)= 3/4(4x-8)
3. solve 3(2x-1)/4= x + 7/2
the 3(2x-1) is over the 4 and the x + 7 is over the 2 i really dont understand them


Sagot :

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
[tex]y = \frac{3}{2} x + 6[/tex]


For the second problem, simply manipulate the equation.
[tex] \frac{1}{3} (6x - 9) = \frac{3}{4} (4x - 8)[/tex]

Distribute
[tex]2x - 3 = 3x - 6[/tex]

Subtract 2x from both sides
[tex]-3 = x - 6[/tex]

Add 6 to both sides
[tex]3 = x[/tex]


For the third problem, do the same thing.
[tex] \frac{3(2x-1)}{4} = x + \frac{7}{2} [/tex]

We must first multiply both sides by 4.
[tex]3(2x-1) = 4x + 14[/tex]

Distribute
[tex]6x - 3 = 4x + 14[/tex]

Subtract 4x from both sides
[tex]2x - 3 = 14[/tex]

Add 3 to both sides
[tex]2x = 17[/tex]

Divide both sides by 2
[tex]x = \frac{17}{2} \ or\ 8 \frac{1}{2} [/tex]