Ask question

Ask question

All categories

Dennis_Churaev [7]

3 years ago

15

2/3y+1=1/6y+8 What does y equal

2 answers:

Andrej [43]3 years ago

8 0

2/3y+1=1/6y+8
Multiply both sides of the equation by 6 you get
4y+6=y+48

4y+6=y+48
Move variable to the left side and change its sign
4y-y+6=48

4y-y=48-6
Collect like terms
3y=48-6

3y=42
Divide both sides of the equation by 3 and your final answer is
y=14

You’re answer is: y=14

joja [24]3 years ago

4 0

Answer:

y = 14

Step-by-step explanation:

4/6y+1 = 1/6y + 8

1/2y = 7

y = 14

Hope this helps!

You might be interested in

Which of the value of x satisfies the equation below -2x-7=12

Masteriza [31]

Answer is A.
You start by adding the 7 to the opposite side. Giving you 19. You divide both sides by -2 which gives you x on one side and -9.5 on the other.

-2x-7=12
-2x=12(+7)
-2x=19
x=19/-2
x=-9.5

8 0

3 years ago

5x-6=10-3x solve for x

hodyreva [135]

Answer:

x = 2

Step-by-step explanation:

Start by combining like terms.

Collecting x terms onto the left side results in 8x;

collecting constants onto the right side results in 16.

Therefore, 8x = 16, and x = 2

4 0

3 years ago

Read 2 more answers

How does the function f(x) = a ln x compare to the parent function when |a| > 1?

VladimirAG [237]

Answer: The graph is stretched.

Step-by-step explanation:

5 0

3 years ago

Steven has some money if he spends nine dollars then he will have 3/5 of the amount he started with

Vesna [10]

Total money * fraction = part
$x* \frac{3}{5} =9 \\ ( \frac{5}{3}) * \frac{3x}{5} = 9* \frac{5}{3} \\ x= \frac{45}{3} =15$

5 0

3 years ago

Read 2 more answers

Which set of parametric equations over the interval 0 ≤ t ≤ 1 defines a line segment with initial point (–5, 3) and terminal poi

Ksivusya [100]

Given:

A line segment with initial point (–5, 3) and terminal point (1, –6).

To find:

The set of parametric equations over the interval 0 ≤ t ≤ 1 which defines the given line segment.

Solution:

Initial point is (–5, 3). So,

$x(0)=-5,y(0)=3$

Terminal point is (1, –6).

$x(1)=1,y(1)=-6$

Check which of the given set of parametric equations satisfy $x(0)=-5,y(0)=3,x(1)=1,y(1)=-6$ .

Put t=1 in each set of parametric equations.

In option A,

$y(1)=3-6(1)=3-6=-3\neq -6$

So, option A is incorrect.

In option B,

$y(1)=1-6(1)=1-6=-5\neq -6$

So, option B is incorrect.

In option C,

$y(1)=3-9(1)=3-9=-6$

$x(1)=-5+6(1)=-5+6=1$

Put t=0, in this set of parametric equations.

$x(0)=-5+6(0)=-5$

$y(0)=3-9(0)=3$

So, option C is correct.

In option D,

$y(1)=1-7(1)=1-7=-3\neq -6$

$x(1)=-5+8(1)=-5+8=3\neq 1$

So, option D is incorrect.

8 0

3 years ago