Question

Can someone please explain how to solve these three algebra problems? Thank you :)

Answer 1

When you divide a variable with an exponent by a variable with an exponent, you subtract the exponents. (You can only combine exponents when the bases/variable is the same, however this rule is a bit different for addition/sub)

For example:   [x²    x is where the base is]

$\frac{x}{y}$ You can't simplify this because they are different variables (x and y)

$\frac{x^5}{x^3} =x^{5-3}=x^2$ You can combine the exponents since they have the same variable (x)

When an exponent is negative, you move the variable and the exponent to the other side of the fraction to make the exponent positive.

For example:

$\frac{x^3}{x^4}=x^{3-4}=x^{-1}=\frac{1}{x^1}$

$\frac{y^2}{y^8} =y^{2-8}=y^{-6}=\frac{1}{y^6}$

$\frac{xy^3}{y^5}=(x)(y^{3-5})=(x)(y^{-2}) = (x)(\frac{1}{y^2})=\frac{x}{y^2}$

#1:

$\frac{14xy-26x^2y^2+16x^3y^3}{4x^2y^2}$ You can separate them if it makes it easier:

$\frac{14xy}{4x^2y^2} -\frac{26x^2y^2}{4x^2y^2}+\frac{16x^3y^3}{4x^2y^2}$

$((\frac{14}{4})(x^{1-2})(y^{1-2}))-((\frac{26}{4})(x^{2-2})(y^{2-2}))+((\frac{16}{4})(x^{3-2})(y^{3-2}))$     When you have an exponent of 0, the result will always equal 1.

Your answer should be:

$\frac{7}{2xy}-\frac{13}{2}+4xy$   [if you have to make only one fraction, make the denominators the same (2xy) to combine them]

#2:

$\frac{x^3-4x^2+2x-3}{x+2}$  I don't think you can simplify this, but I can be wrong, so I'm not sure about this problem

#3

$\frac{x^2+7x-8}{x+8}$     Factor the numerator

$\frac{(x+8)(x-1)}{x+8}$    You can cancel out (x + 8)/(x + 8) bc it = 1

Your answer is:

x - 1