4x+12>4 and 3x-25>-13
Solution {xIx> ? }
I get so lost on these.
Try to get the x by itself, just like a=bx + c, where you would subtract c from both sides then divide both sides by b. since the coefficient for x on both are positive, you don't have to worry about the greater than/less than signs.
so for the first, subtract 12 from both sides then divide both sides by 4. it's the same concept for the other side.
So I got x>2 for the first one and x>4 for the second one, now how does that fit into {xIx> ? }
Shouldn't it be just one number?
4x+12>4 and 3x-25>-13
4x > -8 and 3x > 12
x > -2 and x > 4
so it looks like x > 4
write it in the notation asked for
Ok, great thanks for the help both of you!
To solve the inequalities 4x + 12 > 4 and 3x - 25 > -13, we need to isolate the variable x on one side of the inequalities.
Let's start with the first inequality: 4x + 12 > 4.
1. Subtract 12 from both sides of the inequality:
4x + 12 - 12 > 4 - 12
4x > -8
2. Divide both sides of the inequality by 4 (the coefficient in front of x):
(4x)/4 > (-8)/4
x > -2
Now let's move on to the second inequality: 3x - 25 > -13.
1. Add 25 to both sides of the inequality:
3x - 25 + 25 > -13 + 25
3x > 12
2. Divide both sides of the inequality by 3 (the coefficient in front of x):
(3x)/3 > 12/3
x > 4
Therefore, the solution set for the inequalities is {x | x > -2 and x > 4}.
Now, if you need to write the solution set in interval notation, you need to find the common interval between the two solutions. Since x has to be greater than both -2 and 4, the common interval is x > 4. So the solution set can be written as {x | x > 4}.