What is the standard form of # y= (6x-4)(x+3)-(2x-1)(3x-2)#?

1 Answer
Júlio B. · Tanish J.
Dec 30, 2015

#21x-y=14#

Explanation:

To find the standard form, you have to multiply the content of the parenthesis. First, the first pair:
The first number of the first parenthesis multiplies the numbers in the second one: #6x * x + 6x * 3 = 6x^2 + 18x#. Then we add the multiplication of the second number in the first parenthesis by the numbers in the second one: #-4 * x + (-4) * 3 = -4x -12# and join them
:
#6x^2 +18x -4x -12 = 6x^2 +14x -12#.

Now, just do the same with the second pair:

# 2x * 3x + 2x * (-2) = 6x^2 -4x# and #(-1) * (3x) + (-1) * (-2) = -3x + 2#

And now put them together: # 6x^2 -4x -3x +2 = 6x^2 -7x +2#

And, finally, join the content from the two parenthesis:
#y=6x^2 +14x -12 -(6x^2 -7x +2)=#
#y=6x^2 - 6x^2+14x+7x-12-2 =#
#y=21x -14#

The standard form of a linear equation is #Ax+By=C#

Therefore, we can re-arrange the terms to bring the equation in its standard form as:

#21x-y=14#

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