If RS = 5y + 3, ST = 3y + 7, RT=13y - 30, find the value of y, and find RS, ST, and RT

Answer 1

Given:

1. RS = 5y + 3
2. ST = 3y + 7
3. RT = 13y - 30

Since RS, ST, and RT are segments of a line, we can use the segment addition postulate:

1. RS + ST = RT

Substituting the given expressions:

1. (5y + 3) + (3y + 7) = 13y - 30

Combining like terms:

1. 8y + 10 = 13y - 30

Subtracting 8y from both sides:

1. 10 = 5y - 30

Adding 30 to both sides:

1. 40 = 5y

Dividing both sides by 5:

1. y = 8

Now, we can find the lengths of RS, ST, and RT:

1. RS = 5y + 3 = 5(8) + 3 = 43
2. ST = 3y + 7 = 3(8) + 7 = 31
3. RT = 13y - 30 = 13(8) - 30 = 74

Therefore, y = 8, RS = 43, ST = 31, and RT = 74.

Answer 2

By the line segment addition postulate,

RS + ST = RT

and, substituting the given values into this equation,

[5y+3] + [3y+7] = [13y - 30]

Combining like terms,

8y + 10 = 13y -30

Subtractibg 8y from and adding 40 to both sides,

5y = 40

Dividing by the coefficient of y:

y = 8

RS = 5y+3 = 5*8+3 = 43

ST = 3y+7 = 3*8+7 = 31

RS + ST = 43 + 31 = 74

RT = 13y-30 = 13*8-30 = 104-30 = 74

And this also verifies our computations.