SOLUTION: I need to prove that
((x+4)/(3x^2-7x))=(((1/x)+(4/x^2))/(3-(7/x)))
By showing the steps,
So far I have tried working from the right side of the equation to get to the left
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Question 775978: I need to prove that
((x+4)/(3x^2-7x))=(((1/x)+(4/x^2))/(3-(7/x)))
By showing the steps,
So far I have tried working from the right side of the equation to get to the left, and made it into a complex fraction and combined it, then multiplied the numerator by the denominators reciprocal in order to make it a normal fraction and ended up with ((1-x)(1-4)/(3-7)(x^2-x))
Found 3 solutions by MathLover1, tanjo3, MathTherapy:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
Answer by tanjo3(60) (Show Source): You can put this solution on YOUR website!
Let's leave the left side as is and only deal with the right side:
For the numerator and denominator, find a common denominator:
We can rewrite it as:
Division with fraction means taking the reciprocal of the second number and multiplying.
Simplify.
Now we can see that the right side is equal to the left side.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
I need to prove that
((x+4)/(3x^2-7x))=(((1/x)+(4/x^2))/(3-(7/x)))
By showing the steps,
So far I have tried working from the right side of the equation to get to the left, and made it into a complex fraction and combined it, then multiplied the numerator by the denominators reciprocal in order to make it a normal fraction and ended up with ((1-x)(1-4)/(3-7)(x^2-x))
Simplifying the right-side: , we get:
÷
÷
÷ ----- Multiplying 1st expression by LCD, , and 2nd, by LCD, x
* ------ Changing ÷ to * and inverting DIVISOR
* ------- -------
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f(x)= 7x^2+28x+28
i think i solved it this far.
f(x)=7(x^2+4x)+28
f(x)=7(x^2+4x+4-4)+ (answered by jsmallt9)