Equation1

Algebra

Equation1

Postby Guest » Sat Oct 06, 2018 12:48 am

Find each:

The sum of two numbers is s; m times their sums is equal to n times their difference.

a and c = the numbers.

a + c = s

a + c = ms = a - c = n

Not sure how to continue.
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Re: Equation1

Postby Guest » Tue Oct 09, 2018 3:06 am

[tex]\begin{array}{|l}a+c=s\\m(a+c)=n(a-c)\end{array}[/tex]

[tex]\begin{array}{|l}c=s-a\\ms=n(a-(s-a))=n(2a-s)\end{array}[/tex]

[tex]2a-s=\frac{ms}{n}[/tex]

[tex]2a=s+\frac{ms}{n}=\frac{m+n}{n}\cdot s[/tex]

[tex]a=\frac{m+n}{2n}\cdot s[/tex]

[tex]c=s-a=\left(1-\frac{m+n}{2n}\right)\cdot s=\frac{n-m}{2n}\cdot s[/tex]

Verification:

[tex]a+c=\left(\frac{m+n}{2n}+\frac{n-m}{2n}\right)\cdot s=\frac{2n}{2n}\cdot s=s[/tex]

[tex]m(a+c)-n(a-c)=m.s-\cancel{n}\cdot\frac{\cancel{2}m}{\cancel{2}\cancel{n}}\cdot s=m.s-m.s=0[/tex]

[tex]\Rightarrow m(a+c)=n(a-c)[/tex]

Answer: [tex]a=\frac{m+n}{2n}\cdot s;\ c=\frac{n-m}{2n}\cdot s[/tex]
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Re: Equation1

Postby Guest » Tue Oct 09, 2018 9:52 am

Thanks.
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