Algebra

Algebra

Postby Guest » Wed Apr 04, 2018 3:28 am

I'm new to the forum and Maths is not my subject. I have to prepare for a test which includes a section of Maths, too. So, please help me solve this Word problem.
Thanks
If the numerator of a fraction is increased by 1, the fraction becomes 1. If its numerator is increased by 9 it becomes 2. What is the fraction?
Thanks again.
Guest
 

Re: Algebra

Postby nathi123 » Wed Apr 04, 2018 1:57 pm

If the fraction is [tex]\frac{x}{y}\Rightarrow \frac{x+1}{y}=1\Leftrightarrow x+1=y ,(1)[/tex]
[tex]\frac{x+9}{y}=2 \Leftrightarrow x+9=2y[/tex] (2);
So we have[tex]\begin{array}{|l} x + 1= y \\ x +9=2y \end{array} \Leftrightarrow \begin{array}{|l} x + 9 = 2(x+1) \\ x +1 = y\end{array}[/tex]
[tex]\Leftrightarrow \begin{array}{|l} x+9=2x+2\Leftrightarrow x=7 \\ 7+1 = y \end{array} \Rightarrow \frac{x}{y} =\frac{7}{8}[/tex].

nathi123
 
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Re: Algebra

Postby jjess » Tue May 01, 2018 5:19 pm

From first part of problem:
[tex]\frac{x+1}{y}[/tex] = 1 ----> y = x + 1

Second part:
[tex]\frac{x+9}{y}[/tex] = 2 ----> [tex]\frac{x+9}{x+1}[/tex] = 2 (sub y into denominator from above)

multiply 2 by the denominator:
x + 9 = 2x + 2

solve for x:
7 = x

solve for y by subbing x into equation from first part:
y = 7 + 1 = 8

Answer:
7/8

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