NEED HELP SOLVING 3h/2 = 30 and
3(3+r)= 2r+4
so 3x3=9,3xr=3r = 2r+4
9+3r=2r+4
9-9+4=3r+2r ?????
first one:
3h/2 = 30
multiply both sides by 2
3h = 60
divide both sides by 3
h = 20
second one:
3(3+r)= 2r+4
expand
9 + 3r = 2r + 4
3r - 2r = 4 - 9
r = -5
To solve the first equation, 3h/2 = 30, you can follow these steps:
1. Multiply both sides of the equation by 2 to eliminate the fraction: 2 * (3h/2) = 2 * 30.
This gives you: 3h = 60.
2. Divide both sides of the equation by 3 to isolate the variable h: 3h/3 = 60/3.
This gives you: h = 20.
So, the solution to the first equation is h = 20.
Now let's move on to the second equation, 3(3+r) = 2r + 4.
1. Distribute the 3 to both terms inside the parentheses: 3 * 3 + 3 * r = 2r + 4.
This gives you: 9 + 3r = 2r + 4.
2. Simplify the equation. Start by subtracting 2r from both sides: 9 + 3r - 2r = 2r + 4 - 2r.
This gives you: 9 + r = 4.
3. Subtract 9 from both sides of the equation to isolate the variable r: 9 + r - 9 = 4 - 9.
This gives you: r = -5.
So, the solution to the second equation is r = -5.
Note: It seems there was a mistake in your calculations where you wrote "9-9+4=3r+2r". The correct step should have been "9 + r = 4".