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Readme [11.4K]
3 years ago
15

If RS=3x + 1, ST= 2x - 2, and RT=64 find the value of x

Mathematics
1 answer:
Ulleksa [173]3 years ago
3 0
For this question, we can assume that RS + ST = RT.  So now, we just have to plug in the expressions for these lengths, and solve the equation.

(3x+1) + (2x-2) = 64

Because we are only dealing with one operation, the parentheses aren't necessary.

3x + 1 + 2x - 2 = 64

Next, we should combine like terms on the left side of the equation. We know that 3x + 2x = 5x, and that 1 - 2 = -1.

5x - 1 = 64

The goal of an equation is to get the variable alone.  To do this, we have to get rid of the -1 on the left side of the equation.  So, we are going to add 1 to both sides of the equation, to cancel out the -1 on the left side.

5x = 65

Finally, we are going to divide both sides by 5, as this is the inverse operation of multiplication, which is how the 5 and the x are connected.

x = 13

Therefore, the value of x is 13.
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Answer: The correct option is A.

Step-by-step explanation: We are given a polynomial which is a sum of other 2 polynomials.

We are given the resultant polynomial which is : 8d^5-3c^3d^2+5c^2d^3-4cd^4+9

One of the polynomial which are added up is : 2d^5-c^3d^2+8cd^4+1

Let the other polynomial be 'x'

According to the question:

8d^5-3c^3d^2+5c^2d^3-4cd^4+9=x+(2d^5-c^3d^2+8cd^4+1)

x=8d^5-3c^3d^2+5c^2d^3-4cd^4+9-(2d^5-c^3d^2+8cd^4+1)

Solving the like terms in above equation we get:

x=(8d^5-2d^5)+(-3c^3d^2+c^3d^2)+(5c^2d^3)+(-4cd^4-8cd^4)+(9-1)

x=6d^5-2c^3d^2+5c^2d^3-12cd^4+8

Hence, the correct option is A.

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