Which linear technique would be best to solve this problem? 3x/4 + 6 = -12
And which quadratic equation would be best for -4x^2=0
Thanks for any suggestions.
I'm confused do you want an answer or what to know what technique would be best?
1. I suggest using the common factor and canceling out the bottom number.
Therefore, the common factor is 4. Multuply each one by 4 such as 3x/4 and 6 and also -12. As a result, the 4 crosses out.
--> (4) 3x/4 + (4) 6 = (4) - 12
--> 3x + 24 = /48
--> 3x = - 72
--> x = - 24
2. You have to divide by -4 and x^2 = 0 and then the square root of 0 is always 0.
I would/could have done the work. I just didn't know which technique is best. The name of the technique. Thank you for your help!
To solve the equation 3x/4 + 6 = -12, the best linear technique would be to isolate the variable x by using algebraic operations. Here's how to do it:
1. Start by subtracting 6 from both sides of the equation to move the constant term to the right side:
3x/4 = -12 - 6
3x/4 = -18
2. Next, multiply both sides of the equation by 4 to cancel out the denominator of 3x/4:
4 * (3x/4) = 4 * (-18)
3x = -72
3. Lastly, divide both sides of the equation by 3 to solve for x:
3x / 3 = -72 / 3
x = -24
So the solution to the equation 3x/4 + 6 = -12 is x = -24.
Now, for the quadratic equation -4x^2 = 0, note that this equation can be solved relatively easily since it's already in its simplest form. Here's how to find the solution:
1. Divide both sides of the equation by -4:
(-4x^2) / -4 = 0 / -4
x^2 = 0
2. Take the square root of both sides of the equation:
√(x^2) = √0
x = 0
Therefore, the solution to the equation -4x^2 = 0 is x = 0.
In summary, the linear technique used to solve the equation 3x/4 + 6 = -12 involved isolating the variable x by performing algebraic operations, while for the quadratic equation -4x^2 = 0, the solution was found by dividing by the coefficient and taking the square root.