once again substitution! can anyone help?
x/y-y/3=5/6
x/5-y/4=-13/10
solve by substitution
$\displaystyle \displaystyle\huge\frac{x}{y}-\frac{y}{3}=\frac{5}{6}$
$\displaystyle \displaystyle\huge\frac{x}{5}-\frac{y}{4}=-\frac{13}{10}$
$\displaystyle \displaystyle\huge\frac{x}{y}$ is inconvenient, hence multiply all terms on both sides of the first equation by "y".
$\displaystyle \displaystyle\huge\ x-\frac{y^2}{3}=\frac{5y}{6}$
$\displaystyle \displaystyle\huge\ x=\frac{y^2}{3}+\frac{5y}{6}$
That's one equation for x.
$\displaystyle \displaystyle\huge\frac{x}{5}=\frac{y}{4}-\frac{13}{10}$
$\displaystyle \displaystyle\huge\ x=\frac{5y}{4}-\frac{65}{10}$
x=x, hence equate the two equations in y only.
Subtract them and the answer is zero, so solve the resulting quadratic in y.
same thing as "Archi meade" wrote there, just now with different numbers
multiply firs with 6 and second with 20
$\displaystyle 3x-2y=5 \;\;\;\;\;\;\;\; 1°$
$\displaystyle 4x-5y=-26 \;\;\;\;\;\;\;\; 2°$
now i u can go as you wish
first one multiply with 4 and second with (-3) (and then sum them )
$\displaystyle 12x-8y=20 \;\;\;\;\;\;\;\;1°$
$\displaystyle -12x+15y=78 \;\;\;\;\;\;\;\;2°$
$\displaystyle 7y=98 \Rightarrow y=98/7 = 14 $
rest you know
or u can :
$\displaystyle 6x-2y=5 \Rightarrow 6x=5+2y \Rightarrow x = \frac {2y+5}{6} \;\;\;\;\;\;\;\;1°$
$\displaystyle 4x-5y=-26 \;\;\;\;\;\;\;\; 2°$
and put in second and solve for "y" and so on for x...
P.S. you don't need to do that this way (that's up to you, and with practice u can decide better how and what would you do, so that you have a less work, and to keep it simple... don't complicate it to much )
Edit: corrected that "6" and "20"... i really don't know from where and when did i type that
here's using substitution :
$\displaystyle \displaystyle \frac {x}{2}- \frac {y}{3} = \frac {5}{6} $
$\displaystyle \displaystyle \frac {x}{5}- \frac {y}{4} = -\frac {13}{10} $
first one multiply with (6) and second with (20) and you will get :
$\displaystyle 3x-2y=5$
$\displaystyle 4x-5y=-26$
now from any ( i'll do first ) you can express either "x" or "y" (i'll do "x" )
$\displaystyle \displaystyle 3x-2y=5 \Rightarrow 3x=2y+5 \Rightarrow x= \frac {2y+5}{3} $
now that "x we put in second equation ...
$\displaystyle \displaystyle 4(\frac {2y+5}{3} ) -5y = -26 $
now multiply with 3 :
$\displaystyle \displaystyle 4(2y+5) -15y = -78 \Rightarrow 8y+20-15y = -78 \Rightarrow 7y = 98 \Rightarrow y = 14 $
now that "y" we will put in first equation
$\displaystyle 3x-28 = 5 \Rightarrow 3x = 33 \Rightarrow x=11 $
P.S. what ? now second equation is with -71/10 , where did go -13/10 ?