In simplifying these two binomials, you need to isolate $\displaystyle x$ to one side of the equation. You can first add 4 from the right side to the left side:

$\displaystyle 3x +8 = 6x-4$

$\displaystyle 3x +12 = 6x$

Next you can subtract the $\displaystyle 3x$ from the left side to the right side:

$\displaystyle 12 = 3x$

Finally you can divide each side by 3 to solve for $\displaystyle x$:

$\displaystyle 4=x$

You can double check this answer by plugging 4 into each binomial and confirm that they are equal to one another.

To simplify these two binomials, you need to isolate $\displaystyle z$ on one side of the equation. You first can add 5 from the right to the left side:

$\displaystyle 4z+10 = 6z-5$

$\displaystyle 4z+15 = 6z$

Next you can subtract $\displaystyle 4z$ from the left to the right side:

$\displaystyle 15 = 2z$

Finally, you can isolate $\displaystyle z$ by dividing each side by 2:

$\displaystyle \frac{15}{2} = z$

You can verify this by plugging $\displaystyle \frac{15}{2}$ into each binomial to verify that they are equal to one another.

To solve for $\displaystyle x$, you need to isolate it to one side of the equation. You can subtract the $\displaystyle 6x$ from the right to the left. Then you can add the 6 from the right to the left:

$\displaystyle x^2+2 = 6x-6$

$\displaystyle x^2-6x+2 = -6$

$\displaystyle x^2-6x+8 = 0$

Next, you can factor out this quadratic equation to solve for $\displaystyle x$. You need to determine which factors of 8 add up to negative 6:

$\displaystyle (x \ \ \ \ \ $(x \ \ \ \ \ \) = 0\)

$\displaystyle (x-2)(x-4)=0$

Finally, you set each binomial equal to 0 and solve for $\displaystyle x$:

$\displaystyle x=2 \ \ \ \ \ \ x=4$

$\displaystyle (a+b)^2=a^2+2ab+b^2$

$\displaystyle a=x$

$\displaystyle b=-3$

$\displaystyle x^2+2(-3)(x)+(-3)^2=x^2-6x+9$

32x + 37 = 43x – 29

Add 29 to both sides:

32x + 66 = 43x

Subtract 32x from both sides:

66 = 11x

Divide both sides by 11:

6 = x

When solving for X in terms of Y, we simplify it so that Y is a variable that is used to represent the value of X:

$\displaystyle 7X-21Y=56$

       $\displaystyle +21Y$        $\displaystyle +21Y$

$\displaystyle 7X=21Y+56$

To find the value for X by itself, we then divide both sides by the coefficient of 7:

$\displaystyle \frac{7X}{7}=\frac{56}{7}+\frac{21Y}{7}$

Which gives the correct answer:

$\displaystyle X=8+3Y$

The question is asking for the simplified version of $\displaystyle (8x^2+3x-5)-(-3.5x^2+x+3)$.

Remember the distributive property of multiplication over addition and subtraction:

$\displaystyle ax^2+bx^2 = (a+b)x^2 \mathbf{AND}$  $\displaystyle ax^2-bx^2 = (a-b)x^2$

$\displaystyle 8x^2+3x-5+3.5x^2-x-3$

Combine like terms.

$\displaystyle 8x^2+3.5x^2+3x-x-5-3$

$\displaystyle 11.5x^2+2x-8$

Recall the order of operations: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.

$\displaystyle x \cdot 2 + 7 ^{2}$

$\displaystyle = x \cdot 2 + 49$

$\displaystyle = 2x + 49$

Using the order of opperations, first simplify the exponent.

$\displaystyle x + 2 \cdot 7 ^{2}$

Next, perform the multiplication.

$\displaystyle = x + 2 \cdot 49$

$\displaystyle = x + 98$

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two $\displaystyle x$ terms and one constant are multiplied; find the three products and add them, as follows:

$\displaystyle \left (\underline{ x}+ 0.4\right ) (\underline{x} - 0.2) (3x\underline{-0.7})$

$\displaystyle x \cdot x \cdot (-0.7) = -0.7x^{2}$

$\displaystyle \left (\underline{ x}+ 0.4\right ) (x \underline{-0.2})(\underline{3x}-0.7)$

$\displaystyle x \cdot (-0.2) \cdot 3x= -0.6x^{2}$

$\displaystyle \left (x+ \underline{0.4}\right ) (\underline{x} - 0.2)(\underline{3x}-0.7)$

$\displaystyle 0.4 \cdot x \cdot 3x = 1.2 x^{2}$

Add: $\displaystyle -0.7x^{2}+ (-0.6x^{2})+ 1.2x^{2} = -0.1x^{2}$.

The correct response is $\displaystyle -0.1$.