Can someone please help me solve these two problems and explain them to me? About the one with proportions though... I've always been taught to "cross multiply" when it comes to one proportion, but when there's an equation of proportions where I have to add and subtract them? How do I solve those? The one problem I have no idea how to solve. Full and thorough explanations please :) Thanks! Here's an attachment of the problems:

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cryssatemp cryssatemp · Answer 1 · 2020-01-16T22:24:29+01:00

Answers:

A) [tex]-\frac{11}{4}[/tex]

C) [tex]m_{1}=\frac{Fr^{2}}{G m_{2}}[/tex]

Step-by-step explanation:

Part 1:

We have the followig equation:

[tex]\frac{x-1}{3}-\frac{x+4}{5}=\frac{4x-1}{8}[/tex]

Calculating the least common multiple (l.c.m) in the denominator in the left side of the equation, being l.c.m=15:

[tex]\frac{5(x-1)-3(x+4)}{15}=\frac{4x-1}{8}[/tex]

Solving for the left part of the equation:

[tex]\frac{2x-17}{15}=\frac{4x-1}{8}[/tex]

Operating with cross product:

[tex]8(2x-17)=15(4x-1)[/tex]

Applying the distributive property:

[tex]16x-136=60x-15[/tex]

Isolating [tex]x[/tex]:

[tex]x=-\frac{121}{44}[/tex]

Dividing numerator and denominator by 11:

[tex]x=-\frac{11}{4}[/tex] Hence, the correct option is A

Part 2:

We have the followig equation:

[tex]F=\frac{G m_{1} m_{2}}{r^{2}}[/tex]

Operating with cross product:

[tex]Fr^{2}=G m_{1} m_{2}[/tex]

Isolating [tex]m_{1}[/tex]:

[tex]m_{1}=\frac{Fr^{2}}{G m_{2}}[/tex] Hence, the correct option is C