All categories

3 years ago

Can someone plz explain this

Mathematics

1 answer:

Marat540 [252]3 years ago

6 0

Step 1. Expand

$4x+2+3x-6$

Step 2. Gather like terms.

$(4x+3x)+2-6$

Step 3. Simplify

$7x+2-6$

Step 4. Simplify

$7x-4$

Done! :) Hope this helps!

You might be interested in

3. What is the solu3. What is the solution of system?

White raven [17]

Hello,

Answer A: no solutions

y-5x=4
and y-5x=-3
==>4=-3 ==> no solutions

5 0

3 years ago

(3x^3+4x^2)+(3x^3-4x^2-9x)

zysi [14]

Answer:3x • (2x2 - 3)

Step-by-step explanation:

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2". 3 more similar replacement(s).

STEP

Equation at the end of step 1

((3•(x3))+(4•(x2)))+(((3•(x3))-22x2)-9x)

STEP

Equation at the end of step

((3•(x3))+(4•(x2)))+((3x3-22x2)-9x)

STEP

Equation at the end of step

((3 • (x3)) + 22x2) + (3x3 - 4x2 - 9x)

STEP

Equation at the end of step

(3x3 + 22x2) + (3x3 - 4x2 - 9x)

STEP

Pulling out like terms

6.1 Pull out like factors :

6x3 - 9x = 3x • (2x2 - 3)

Trying to factor as a Difference of Squares:

6.2 Factoring: 2x2 - 3

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 2 is not a square !!

Ruling : Binomial can not be factored as the

difference of two perfect squares

Final result :

3x • (2x2 - 3)

5 0

3 years ago

(x/x+3)-(8/x+2)=(-13x-6)/(x^2+5x+6)

Fed [463]

Answer:

x = -9

x = 2

Step-by-step explanation:

Given:

$\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{x^2+5x+6}$

First, note that

$x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)$

Since $x+2$ and $x+3$ stay in the denominator, then

$x\neq -2\ \text{and}\ x\neq -3$

Now add two fractions which stay in the left part. The common denominator is $(x+2)(x+3),$ so multiply the numerator of the first fraction by $(x+2)$ and the numerator of the second fraction by $(x+3)$ and subtract them in the numerator:

$\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\\dfrac{x(x+2)-8(x+3)}{(x+2)(x+3)}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\x(x+2)-8(x+3)=-13x-6\\ \\x^2+2x-8x-24=-13x-6\\ \\x^2+2x-8x+13x-24+6=0\\ \\x^2+7x-18=0\\ \\x^2+9x-2x-18=0\\ \\x(x+9)-2(x+9)=0\\ \\(x+9)(x-2)=0\\ \\x+9=0\ \text{or}\ x-2=0\\ \\x=-9\ \text{or}\ x=2$