\(\displaystyle (-x+10)(3x^{2}-4)\)

Use the FOIL method to multiply the binomials given.

F: \(\displaystyle -x(3x^{2})=-3x^{3}\)

O: \(\displaystyle -x(-4)=4x\)

I: \(\displaystyle 10(3x^{2})=30x^{2}\)

L: \(\displaystyle 10(-4)=-40\)

Group any like terms (none for this problem) when putting all the terms back together.

\(\displaystyle -3x^{3}+30x^{2}+4x-40\)

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.

\(\displaystyle = x + 2 \cdot 49\)

\(\displaystyle = x + 98\)