What is the solution of x minus 16 equals 1?
17 answers
The solution to x minus 16 equals 1 is x equals 17.
What is the solution of x minus 16 equals 1? show all work
To find the solution, we need to isolate the variable x.
Given:
x - 16 = 1
Step 1: Add 16 to both sides of the equation.
x - 16 + 16 = 1 + 16
x = 17
Therefore, the solution is x = 17.
Given:
x - 16 = 1
Step 1: Add 16 to both sides of the equation.
x - 16 + 16 = 1 + 16
x = 17
Therefore, the solution is x = 17.
Solve the equation.
18 minus left-parenthesis x plus 8 right-parenthesis equals 7 x minus 6
show all work
18 minus left-parenthesis x plus 8 right-parenthesis equals 7 x minus 6
show all work
To solve the equation, we will simplify both sides and isolate the variable x.
Given: 18 - (x + 8) = 7x - 6
Step 1: Distribute the negative sign to the terms inside the parentheses.
18 - x - 8 = 7x - 6
Step 2: Combine like terms.
10 - x = 7x - 6
Step 3: Move all terms with x to one side of the equation.
-x - 7x = -6 - 10
-8x = -16
Step 4: Divide both sides by -8 to solve for x.
(-8x)/-8 = (-16)/-8
x = 2
Therefore, the solution to the equation 18 - (x + 8) = 7x - 6 is x = 2.
Given: 18 - (x + 8) = 7x - 6
Step 1: Distribute the negative sign to the terms inside the parentheses.
18 - x - 8 = 7x - 6
Step 2: Combine like terms.
10 - x = 7x - 6
Step 3: Move all terms with x to one side of the equation.
-x - 7x = -6 - 10
-8x = -16
Step 4: Divide both sides by -8 to solve for x.
(-8x)/-8 = (-16)/-8
x = 2
Therefore, the solution to the equation 18 - (x + 8) = 7x - 6 is x = 2.
A new outdoor recreation center is being built in Hadleyville. The perimeter of the rectangular playing field is 316 yards. The length of the field is 4 yards less than twice its width. What are the dimensions of the playing field?
show all work
show all work
Let's assume the width of the playing field is x yards.
According to the given information, the length of the field is 4 yards less than twice its width, so the length would be (2x - 4) yards.
The perimeter of a rectangle is given by the formula P = 2l + 2w, where P is the perimeter, l is the length, and w is the width.
Plugging in the given values, we have:
316 = 2(2x - 4) + 2x
Simplifying the equation:
316 = 4x - 8 + 2x
316 = 6x - 8
Add 8 to both sides:
316 + 8 = 6x
324 = 6x
Divide both sides by 6:
(324/6) = x
54 = x
Therefore, the width of the playing field is 54 yards.
To find the length, we can substitute the value of x into the equation (2x - 4):
Length = 2(54) - 4
Length = 108 - 4
Length = 104
Therefore, the dimensions of the playing field are a width of 54 yards and a length of 104 yards.
According to the given information, the length of the field is 4 yards less than twice its width, so the length would be (2x - 4) yards.
The perimeter of a rectangle is given by the formula P = 2l + 2w, where P is the perimeter, l is the length, and w is the width.
Plugging in the given values, we have:
316 = 2(2x - 4) + 2x
Simplifying the equation:
316 = 4x - 8 + 2x
316 = 6x - 8
Add 8 to both sides:
316 + 8 = 6x
324 = 6x
Divide both sides by 6:
(324/6) = x
54 = x
Therefore, the width of the playing field is 54 yards.
To find the length, we can substitute the value of x into the equation (2x - 4):
Length = 2(54) - 4
Length = 108 - 4
Length = 104
Therefore, the dimensions of the playing field are a width of 54 yards and a length of 104 yards.
Is the equation always, sometimes, or never true?
5 x minus 16 plus 2 x equals 2 plus 7 x minus 17
5 x minus 16 plus 2 x equals 2 plus 7 x minus 17
The equation in question is: 5x - 16 + 2x = 2 + 7x - 17.
To determine if the equation is always, sometimes, or never true, we need to solve it and verify.
Combining like terms:
7x - 16 = -15 + 7x
Since we still have an "x" term on both sides of the equation, we can subtract 7x from both sides to eliminate it:
7x - 7x - 16 = -15 - 7x
Simplifying:
-16 = -15 - 7x
Next, we can add 15 to both sides to isolate the variable:
-16 + 15 = -15 + 15 - 7x
Simplifying:
-1 = -7x
Finally, we divide both sides by -7 to solve for x:
-1 / -7 = (-7x) / -7
1/7 = x
So the solution to the equation is x = 1/7.
Therefore, the equation 5x - 16 + 2x = 2 + 7x - 17 is only sometimes true. It is true when x = 1/7, but not for any other values of x.
To determine if the equation is always, sometimes, or never true, we need to solve it and verify.
Combining like terms:
7x - 16 = -15 + 7x
Since we still have an "x" term on both sides of the equation, we can subtract 7x from both sides to eliminate it:
7x - 7x - 16 = -15 - 7x
Simplifying:
-16 = -15 - 7x
Next, we can add 15 to both sides to isolate the variable:
-16 + 15 = -15 + 15 - 7x
Simplifying:
-1 = -7x
Finally, we divide both sides by -7 to solve for x:
-1 / -7 = (-7x) / -7
1/7 = x
So the solution to the equation is x = 1/7.
Therefore, the equation 5x - 16 + 2x = 2 + 7x - 17 is only sometimes true. It is true when x = 1/7, but not for any other values of x.
Solve for h.
upper A equals start fraction 5 over 4 end fraction left-parenthesis h minus 71 right-parenthesis
upper A equals start fraction 5 over 4 end fraction left-parenthesis h minus 71 right-parenthesis
To solve for h, we need to isolate the variable h in the equation:
A = (5/4) * (h - 71)
First, we can divide both sides of the equation by (5/4) to remove it from the left side:
A / (5/4) = h - 71
To simplify, we can multiply the numerator and denominator of the fraction A / (5/4) by the reciprocal of 5/4, which is 4/5:
A * (4/5) = (h - 71) * (4/5)
Next, we can distribute the (4/5) on the right side:
A * (4/5) = (4/5) * (h - 71)
Simplifying:
(4A/5) = (4/5) * (h - 71)
The (4/5) cancels out on both sides, leaving:
4A = h - 71
Now, to isolate h, we can add 71 to both sides:
4A + 71 = h
Therefore, the solution for h is h = 4A + 71.
A = (5/4) * (h - 71)
First, we can divide both sides of the equation by (5/4) to remove it from the left side:
A / (5/4) = h - 71
To simplify, we can multiply the numerator and denominator of the fraction A / (5/4) by the reciprocal of 5/4, which is 4/5:
A * (4/5) = (h - 71) * (4/5)
Next, we can distribute the (4/5) on the right side:
A * (4/5) = (4/5) * (h - 71)
Simplifying:
(4A/5) = (4/5) * (h - 71)
The (4/5) cancels out on both sides, leaving:
4A = h - 71
Now, to isolate h, we can add 71 to both sides:
4A + 71 = h
Therefore, the solution for h is h = 4A + 71.
Write the inequality that represents the following sentence:
Four less than a number is greater than 58.
Four less than a number is greater than 58.
The inequality that represents the sentence "Four less than a number is greater than 58" is:
x - 4 > 58
Where x represents the number.
x - 4 > 58
Where x represents the number.
Solve the compound inequality 1 plus k greater-than-or-equal-to 4 or 5 plus k less-than 2. Explain how the solution set is shown on a number line.
To solve the compound inequality 1 + k ≥ 4 or 5 + k < 2, we will solve each inequality separately and then combine the solution sets.
First, let's solve the inequality 1 + k ≥ 4:
Subtract 1 from both sides:
k ≥ 4 - 1
k ≥ 3
The solution for this inequality is k is greater than or equal to 3.
Now, let's solve the inequality 5 + k < 2:
Subtract 5 from both sides:
k < 2 - 5
k < -3
The solution for this inequality is k is less than -3.
To represent the solution set on a number line, we can plot two points: one at 3 (closed circle since the inequality is ≥), and another at -3 (open circle since the inequality is <).
Then, we shade the number line between these two points to show the values that satisfy the inequality.
So, the solution set on a number line would look like this:
<--- (-∞)--------------•------------------•--------------(∞) --->
-3 3
The shaded region indicates the values of k that satisfy the compound inequality 1 + k ≥ 4 or 5 + k < 2. It includes all real numbers less than -3 and greater than or equal to 3.
First, let's solve the inequality 1 + k ≥ 4:
Subtract 1 from both sides:
k ≥ 4 - 1
k ≥ 3
The solution for this inequality is k is greater than or equal to 3.
Now, let's solve the inequality 5 + k < 2:
Subtract 5 from both sides:
k < 2 - 5
k < -3
The solution for this inequality is k is less than -3.
To represent the solution set on a number line, we can plot two points: one at 3 (closed circle since the inequality is ≥), and another at -3 (open circle since the inequality is <).
Then, we shade the number line between these two points to show the values that satisfy the inequality.
So, the solution set on a number line would look like this:
<--- (-∞)--------------•------------------•--------------(∞) --->
-3 3
The shaded region indicates the values of k that satisfy the compound inequality 1 + k ≥ 4 or 5 + k < 2. It includes all real numbers less than -3 and greater than or equal to 3.
Solve the equation start absolute value 5 x minus 5 end absolute value equals 15. Explain how the solution set is shown on a number line.
To solve the equation, we need to isolate the absolute value expression and remove the absolute value bars.
Start by setting up two separate equations, one with the positive value inside the absolute value and one with the negative value:
1) 5x - 5 = 15
2) 5x - 5 = -15
Solving equation 1:
Add 5 to both sides:
5x - 5 + 5 = 15 + 5
5x = 20
Divide both sides by 5:
(5x)/5 = 20/5
x = 4
Solving equation 2:
Add 5 to both sides:
5x - 5 + 5 = -15 + 5
5x = -10
Divide both sides by 5:
(5x)/5 = -10/5
x = -2
Therefore, the solutions to the equation are x = 4 and x = -2.
Now, let's show the solution set on a number line:
<--- (-∞)---------------------•-----(-2)-----•------(4)-------------------(∞) --->
We place a closed circle at x = 4 and an open circle at x = -2, indicating that these values are included in the solution set.
The shaded region between -2 and 4 represents all the values that satisfy the equation. It includes -2 and 4, as well as any value between them.
Start by setting up two separate equations, one with the positive value inside the absolute value and one with the negative value:
1) 5x - 5 = 15
2) 5x - 5 = -15
Solving equation 1:
Add 5 to both sides:
5x - 5 + 5 = 15 + 5
5x = 20
Divide both sides by 5:
(5x)/5 = 20/5
x = 4
Solving equation 2:
Add 5 to both sides:
5x - 5 + 5 = -15 + 5
5x = -10
Divide both sides by 5:
(5x)/5 = -10/5
x = -2
Therefore, the solutions to the equation are x = 4 and x = -2.
Now, let's show the solution set on a number line:
<--- (-∞)---------------------•-----(-2)-----•------(4)-------------------(∞) --->
We place a closed circle at x = 4 and an open circle at x = -2, indicating that these values are included in the solution set.
The shaded region between -2 and 4 represents all the values that satisfy the equation. It includes -2 and 4, as well as any value between them.
1 answer