SOLUTION: Hi, Can someone assist me with the following problem? x-5/x+4=2/7

Algebra.Com
Question 125879:

Hi, Can someone assist me with the following problem? x-5/x+4=2/7

Found 2 solutions by jim_thompson5910, checkley71:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Start with the given equation


Cross multiply


Distribute



Add 35 to both sides


Subtract 2x from both sides


Combine like terms on the left side


Combine like terms on the right side


Divide both sides by 5 to isolate x




--------------------------------------------------------------
Answer:
So our answer is (which is in decimal form)

Answer by checkley71(8403)   (Show Source): You can put this solution on YOUR website!
(X-5)/(X+4)=2/7 IF THIS IS THE EQUATION THEN WE HAVE TO CROSS MULTIPLY.
7(X-5)=2(X+4)
7X-35=2X+8
7X-2X=8+35
5X=43
X=43/5 ANSWER.
------------------------------------------
X-(5/X)+4=2/7 IF THIS IS THE EQUATION THEN:
(X^2-5)/X+4=2/7
(X^2-5)/X=2/7-4
(X^2-5)/X=(2-4*7)/7
(X^2-5)/X=(2-28)/7
(X^2-5)/X=-26/7 NOW CROSS MULTIPLY.
7(X^2-5)=-26X
7X^2-35+26X=0
7X^2+26X-35=0USING THE QUADRATIC EQUATION WE GET;

X=1.05 & -4.76





RELATED QUESTIONS

Hi, Can someone assist me ont the following problem?... (answered by jim_thompson5910)
Hi, Can someone assist me with the following problem? Find the LCM... (answered by checkley71)
ok can you assist me with this problem i have got stuck.... (answered by Flannery)
Could you please assist me with this problem? y=-5 x 3 +... (answered by stanbon)
Hi, can someone please help me solve the following inequality? 2 < (5 - 2x)/3 < 7 (answered by funmath)
Can someone assist me? Thank You Find and label the vertex... (answered by checkley71)
Can someone help me on this problem? I need to evaluate the following expression: (answered by stanbon,Osran_Shri,pwac)
Hi can you help me with the following problem? Write the answer in standard form of... (answered by checkley71)
Hi, can someone please help me solve the following... x^2 + y^2 = 14, x^2 - y^2 =... (answered by stanbon)