PROBLEM ONE
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Solving for x in 2x + 5y > -1.
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Step 1 ) Subtract 5y from both sides.
2x + 5y > -1
2x + 5y - 5y > -1 - 5y
2x > -1 - 5y
Step 2 ) Divide both sides by 2.
2x > -1 - 5y
![\displaystyle(2x)/(2) > \displaystyle(-1 - 5y)/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/lmthq8n5m1zc83ybmyf7pf7ntzx3soab0q.png)
![\displaystyle\ x > (-1 - 5y)/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/bl8vqsc6rc7l8uoimfkz8v6vc24r4ndgmp.png)
So, the solution for x in 2x + 5y > -1 is...
![\displaystyle\ x > (-1 - 5y)/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/bl8vqsc6rc7l8uoimfkz8v6vc24r4ndgmp.png)
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Solving for y in 2x + 5y > -1.
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Step 1 ) Subtract 2x from both sides.
2x + 5y > -1
2x - 2x + 5y > -1 - 2x
5y > -1 - 1x
Step 2 ) Divide both sides by 5.
5y > -1 - 1x
![\displaystyle(5x)/(5) > (-1 -1x)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/51ax2v6d2vgrtwh6xrn672iw9anbo2tmq8.png)
![\displaystyle\ x > (-1 -1x)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/k8td6orpus8grtvtobzvkabb1i19idps7b.png)
So, the solution for y in 2x + 5y > -1 is...
![\displaystyle\ x > (-1 -1x)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/k8td6orpus8grtvtobzvkabb1i19idps7b.png)
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PROBLEM TWO
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Solving for x in 4x - 3 < -3.
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Step 1 ) Subtract 3 from both sides.
4x - 3 < -3
4x -3 - 3 < -3 - 3
4x < 0
Step 2 ) Divide both sides by x.
4x < 0
![\displaystyle(4x)/(4) <(0)/(4)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/31gohyg6y784epr2rlf0c3z9gzgumgpdd0.png)
x < 0
So, the solution for x in 4x - 3 < -3 is...
x < 0
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- Marlon Nunez