SOLUTION: My name is dylan, I have 2 problems I cannot solve. I am in 7th grade in Pre-Algebra. This is solving multi-step inequalities 1) -0.59 < (t/-4) -0.09 2) (z+5)/2

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Question 275353: My name is dylan, I have 2 problems I cannot solve. I am in 7th grade in Pre-Algebra.
This is solving multi-step inequalities
1) -0.59 < (t/-4) -0.09
2) (z+5)/2 < (4-z)/7

Found 2 solutions by stanbon, dabanfield:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
solving multi-step inequalities
1) -0.59 < (t/-4) -0.09
Multiply the left and the right side by 100 to
get rid of the denominators:
---
-59 < -25t - 9
Add 25t to the left side and to the right side to get:
25t -59 < -9
---
Add 59 to the left side and to the right side to get:
25t < 50
---
Divide both sides by 25 to solve for "t":
t < 2
=======================================================
2) (z+5)/2 < (4-z)/7
Multiply both sides 14 to get:
7(z+5) < 2(4-z)
---
Use the distributive law on both sides of the inequality to get:
7z + 35 < 8 - 2z
Add 2z to both sides to get:
9z + 35 < 8
Subtract 35 from both sides to get:
9z < -27
Divide both sides by 9 to get:
z < -3
======================================
Good luck with you studies.
Cheers,
Stan H.

Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website!
1) -0.59 < (t/-4) -0.09
2) (z+5)/2 < (4-z)/7
1.) -.59 < (t/-4) - .09
Add .09 to both sides:
-.50 < t/-4
Multiply both sides by -4. When multiplying by a negative the direction of the inequality is reversed:
-4*-.5 > t*(-4/-4)
2 > t
2.) (z+5)/2 < (4-z)/7
Multiply both sides first by 2 and then by 7 to remove the fractions:
(z+5)*(2/2) < 2*(4-z)/7
7*(z+5) < 2*(4-z)*(7/7)
7(z+5) < 2*(4-z)
7z + 35 < 8 - 2z
7z+2z < 8 - 35
9z < -27
z < -3

SOLUTION: My name is dylan, I have 2 problems I cannot solve. I am in 7th grade in Pre-Algebra. This is solving multi-step inequalities 1) -0.59 < (t/-4) -0.09 2) (z+5)/2 < (4-z)/