Solve: 6(y-5) = 2(10 + 3y)

2 answers:

5 0

Simplifying
6[y + -5] = 2(10 + 3y)

Reorder the terms:
6[-5 + y] = 2(10 + 3y)
[-5 * 6 + y * 6] = 2(10 + 3y)
[-30 + 6y] = 2(10 + 3y)
-30 + 6y = (10 * 2 + 3y * 2)
-30 + 6y = (20 + 6y)

Add '-6y' to each side of the equation.
-30 + 6y + -6y = 20 + 6y + -6y

Combine like terms: 6y + -6y = 0
-30 + 0 = 20 + 6y + -6y
-30 = 20 + 6y + -6y

Combine like terms: 6y + -6y = 0
-30 = 20 + 0
-30 = 20

Solving
-30 = 20

Couldn't find a variable to solve for.

no solution

topjm [15]3 years ago

4 0

Answer:

no solution

Step-by-step explanation:

simplifying

6(y + -5) = 2(10 + 3y)

reorder the terms:

6(-5 + y) = 2(10 + 3y)

(-5 * 6 + y * 6) = 2(10 + 3y)

(-30 + 6y) = 2(10 + 3y)

-30 + 6y = (10 * 2 + 3y * 2)

-30 + 6y = (20 + 6y)

add '-6y' to each side of the equation.

-30 + 6y + -6y = 20 + 6y + -6y

combine like terms: 6y + -6y = 0

-30 + 0 = 20 + 6y + -6y

-30 = 20 + 6y + -6y

combine like terms: 6y + -6y = 0

-30 = 20 + 0

-30 = 20

solving for:

-30 = 20

the left and right sides are not equal, so there isn't any solution!

You might be interested in

Determine whether the following problem involves a permutation or a combination and explain your answer. How many different -le

Alex_Xolod [135]

Answer:

B. The problem involves a combination because the order in which the letters are selected does not matter.

Step-by-step explanation:

Computation technique is a method of statistics to find possible ways of combination. In computation technique it is assumed that order does not matter and letters will be selected at random. Permutation is a statistics technique to find possible ways of combination when the order does matter. Permutation technique cannot be used when order does not matter.

4 0

3 years ago

The volume of a cone is 565.2 cubic inches. The height of the cone is 15 inches. What is the radius of the cone, rounded to the

Ludmilka [50]

I
put
6
and
got
1
point

(aka right)

5 0

3 years ago

Solving equations (algebra). Thank you!

LenKa [72]

Answer:

$\large\boxed{\dfrac{1}{x^2}+x^2=23}$

Step-by-step explanation:

$\left(\dfrac{1}{x}+x\right)^2=25\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\\left(\dfrac{1}{x}\right)^2+2(x\!\!\!\!\diagup)\left(\dfrac{1}{x\!\!\!\!\diagup}\right)+x^2=25\\\\\dfrac{1}{x^2}+2+x^2=25\qquad\text{subtract 2 from both sides}\\\\\dfrac{1}{x^2}+x^2=23$