Question 178562
</pre><font size=4><b>
Since no parenthesis used, we assumed the following:
{{{(x/3)=4/(x+1)}}}
cross multiply,
{{{x(x+1)=4*3}}}
{{{x^2+x=12}}}--->{{{x^2+x-12=0}}}
By QUADRATIC, where{{{system(a=1,b=1,c=-12)}}}
{{{x=(-b+-sqrt(b^2-4ac))/(2a)=(-1+-sqrt(1^2-4*(1)(-12)))/(2*1)}}}
{{{x=(-1+-sqrt(1+48))/2=(-1+-sqrt(49))/2=(1+-7)/2}}}
2 values:
{{{x=(-1+7)/2=6/2=highlight(3)}}} Answer
{{{x=(-1-7)/2=-8/2=-4}}}, discard
Check:
{{{3/3=4/(3+1)}}}
{{{1=4/4}}}
{{{1=1}}}
Thank you,
Jojo</pre>