To solve the equation, start by simplifying both sides of the equation.
On the left side, distribute the 1/4 to 8 and -16g:
1/4(8-16g) = (1/4) * 8 + (1/4) * (-16g) = 2 - 4g
So the equation becomes:
2 - 4g + 8 = g
Combine like terms:
10 - 4g = g
To isolate the variable g, add 4g to both sides:
10 = 5g
Finally, divide both sides by 5 to solve for g:
10/5 = g
2 = g
Therefore, the solution to the equation 1/4(8-16g) + 8 = g is g = 2.
Using the properties of equality, solve the equation 1/4(8−16g)+8=g .
5 answers
Shyr solved the equation w/2−7=11 . She seems to have made an error in her work because the solution she found does not work. Observe Sehyr’s work to find her error and correct it to find the solution to the equation. Sehyr’s work: w/2−7=11 w/2−7+7=11+7 Using the Addition Property of Equality w/2/2=18/2 Using the Division Property of Equality
To correct Sehyr's work, the mistake is made when adding 7 on both sides of the equation. Let's correct it properly:
Starting with the equation: w/2 - 7 = 11
Add 7 to both sides of the equation:
w/2 - 7 + 7 = 11 + 7
w/2 = 18
To solve for w, we need to multiply both sides of the equation by 2 to eliminate the fraction:
2(w/2) = 2(18)
w = 36
Therefore, the correct solution to the equation w/2 - 7 = 11 is w = 36.
Starting with the equation: w/2 - 7 = 11
Add 7 to both sides of the equation:
w/2 - 7 + 7 = 11 + 7
w/2 = 18
To solve for w, we need to multiply both sides of the equation by 2 to eliminate the fraction:
2(w/2) = 2(18)
w = 36
Therefore, the correct solution to the equation w/2 - 7 = 11 is w = 36.
How many of the following equations have the solution set {5} ? 2(x−3)+1=5 3b/15=1 2h−30=40
To determine which equations have the solution set {5}, we can substitute 5 into each equation and check if it satisfies the equation.
1) 2(x - 3) + 1 = 5:
Substituting x = 5:
2(5 - 3) + 1 = 5
2(2) + 1 = 5
4 + 1 = 5
5 = 5
Therefore, the equation 2(x - 3) + 1 = 5 has the solution set {5}.
2) 3b/15 = 1:
Substituting b = 5:
3(5)/15 = 1
15/15 = 1
1 = 1
Therefore, the equation 3b/15 = 1 has the solution set {5}.
3) 2h - 30 = 40:
Substituting h = 5:
2(5) - 30 = 40
10 - 30 = 40
-20 = 40
Since -20 is not equal to 40, the equation 2h - 30 = 40 does not have the solution set {5}.
Hence, out of the three given equations, only 2(x - 3) + 1 = 5 and 3b/15 = 1 have the solution set {5}.
1) 2(x - 3) + 1 = 5:
Substituting x = 5:
2(5 - 3) + 1 = 5
2(2) + 1 = 5
4 + 1 = 5
5 = 5
Therefore, the equation 2(x - 3) + 1 = 5 has the solution set {5}.
2) 3b/15 = 1:
Substituting b = 5:
3(5)/15 = 1
15/15 = 1
1 = 1
Therefore, the equation 3b/15 = 1 has the solution set {5}.
3) 2h - 30 = 40:
Substituting h = 5:
2(5) - 30 = 40
10 - 30 = 40
-20 = 40
Since -20 is not equal to 40, the equation 2h - 30 = 40 does not have the solution set {5}.
Hence, out of the three given equations, only 2(x - 3) + 1 = 5 and 3b/15 = 1 have the solution set {5}.
1 answer