−2z+4=19

11 answers

To solve for z, we need to isolate it on one side of the equation.

Starting with the equation −2z + 4 = 19, we can begin by subtracting 4 from both sides:

−2z + 4 - 4 = 19 - 4
−2z = 15

Next, divide both sides of the equation by -2:

−2z/-2 = 15/-2
z = -7.5

Therefore, the solution to the equation −2z + 4 = 19 is z = -7.5.
4(x−6)=9
To solve for x, we need to isolate it on one side of the equation.

Starting with the equation 4(x - 6) = 9, we can begin by distributing the 4 to the terms inside the parentheses:

4x - 24 = 9

Next, we can add 24 to both sides of the equation to move the constant term to the right side:

4x - 24 + 24 = 9 + 24
4x = 33

Finally, divide both sides of the equation by 4 to solve for x:

4x/4 = 33/4
x = 33/4

Therefore, the solution to the equation 4(x - 6) = 9 is x = 33/4.
−2(7x+11)=15
To solve for x, we need to isolate it on one side of the equation.

Starting with the equation −2(7x + 11) = 15, we can simplify the left side by distributing the -2 to the terms inside the parentheses:

-14x - 22 = 15

Next, we can add 22 to both sides of the equation to move the constant term to the right side:

-14x - 22 + 22 = 15 + 22
-14x = 37

Finally, divide both sides of the equation by -14 to solve for x:

-14x / -14 = 37 / -14
x = -37/14

Therefore, the solution to the equation −2(7x + 11) = 15 is x = -37/14.
10y+19−2y=−125
To solve for y, we need to isolate it on one side of the equation.

Starting with the equation 10y + 19 - 2y = -125, we can simplify the left side of the equation by combining like terms:

10y - 2y + 19 = -125

This simplifies to:

8y + 19 = -125

Next, we can subtract 19 from both sides to move the constant term to the right side:

8y + 19 - 19 = -125 - 19
8y = -144

Finally, divide both sides of the equation by 8 to solve for y:

(8y) / 8 = (-144) / 8
y = -18

Therefore, the solution to the equation 10y + 19 - 2y = -125 is y = -18.
A movie theater offers two options for membership. The first option costs $15 per ticket but an initial fee of $10. The second option costs $12.50 per ticket but an initial fee of $40 . After how many tickets are both options the same price?

____ tickets
Let's represent the number of tickets as "x".

For the first option, the cost is $15 per ticket plus an initial fee of $10. The total cost will be (15x + 10).

For the second option, the cost is $12.50 per ticket plus an initial fee of $40. The total cost will be (12.50x + 40).

To find out when both options are the same price, we need to set up an equation and solve for x:

15x + 10 = 12.50x + 40

Subtracting 12.50x from both sides:

15x - 12.50x + 10 = 12.50x - 12.50x + 40

2.5x + 10 = 40

Subtracting 10 from both sides:

2.5x + 10 - 10 = 40 - 10

2.5x = 30

Dividing both sides by 2.5 to solve for x:

(2.5x) / 2.5 = 30 / 2.5

x = 12

Therefore, after purchasing 12 tickets, both options will be the same price
There are two different vats. The first vat has 12 gallons is being filled at a rate of 3 gallons per minute. The second vat has 120 gallons and is leaking fluid at a rate of 7 gallons per minute. After how many seconds will both vats have the same amount of liquid?

____ minutes
Let's represent the number of minutes as "m".

For the first vat, it is being filled at a rate of 3 gallons per minute. The total amount of liquid in the first vat can be represented as 12 + 3m.

For the second vat, it is leaking fluid at a rate of 7 gallons per minute. The total amount of liquid in the second vat can be represented as 120 - 7m.

To find out when both vats have the same amount of liquid, we need to set up an equation and solve for m:

12 + 3m = 120 - 7m

Adding 7m to both sides:

12 + 3m + 7m = 120 - 7m + 7m

10m + 12 = 120

Subtracting 12 from both sides:

10m + 12 - 12 = 120 - 12

10m = 108

Dividing both sides by 10 to solve for m:

(10m) / 10 = 108 / 10

m = 10.8

Therefore, after 10.8 minutes, both vats will have the same amount of liquid.

To convert this to seconds, we multiply by 60:

10.8 minutes * 60 seconds/minute = 648 seconds

Therefore, after 648 seconds, both vats will have the same amount of liquid.