solve by substitution

**mafai44** · August 12th 2010, 11:32 AM

once again substitution! can anyone help?

x/y-y/3=5/6
x/5-y/4=-13/10

solve by substitution

**Archie Meade** · August 12th 2010, 11:53 AM

Originally Posted by mafai44

once again substitution! can anyone help?

x/y-y/3=5/6
x/5-y/4=-13/10

solve by substitution

$\displaystyle\huge\frac{x}{y}-\frac{y}{3}=\frac{5}{6}$

$\displaystyle\huge\frac{x}{5}-\frac{y}{4}=-\frac{13}{10}$

$\displaystyle\huge\frac{x}{y}$ is inconvenient, hence multiply all terms on both sides of the first equation by "y".

$\displaystyle\huge\ x-\frac{y^2}{3}=\frac{5y}{6}$

$\displaystyle\huge\ x=\frac{y^2}{3}+\frac{5y}{6}$

That's one equation for x.

$\displaystyle\huge\frac{x}{5}=\frac{y}{4}-\frac{13}{10}$

$\displaystyle\huge\ x=\frac{5y}{4}-\frac{65}{10}$

x=x, hence equate the two equations in y only.
Subtract them and the answer is zero, so solve the resulting quadratic in y.

**mafai44** · August 12th 2010, 12:12 PM

so sorry lol i wrote the equation wrong x/2-y/3=5/6
x/5-y/4=-13/10

**yeKciM** · August 12th 2010, 12:25 PM

Originally Posted by mafai44

so sorry lol i wrote the equation wrong x/2-y/3=5/6
x/5-y/4=-13/10

same thing

as "Archi meade" wrote there, just now with different numbers

multiply firs with 6 and second with 20

$3x-2y=5 \;\;\;\;\;\;\;\; 1°$
$4x-5y=-26 \;\;\;\;\;\;\;\; 2°$

now i u can go as you wish

first one multiply with 4 and second with (-3) (and then sum them )

$12x-8y=20 \;\;\;\;\;\;\;\;1°$
$-12x+15y=78 \;\;\;\;\;\;\;\;2°$

$7y=98 \Rightarrow y=98/7 = 14$

rest you know

or u can :
$6x-2y=5 \Rightarrow 6x=5+2y \Rightarrow x = \frac {2y+5}{6} \;\;\;\;\;\;\;\;1°$
$4x-5y=-26 \;\;\;\;\;\;\;\; 2°$

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

**Archie Meade** · August 12th 2010, 01:10 PM

Typo yeKciM !

3x-2y=5

**yeKciM** · August 12th 2010, 01:16 PM

aaaaah

thank you , i'll fix it now

**mafai44** · August 12th 2010, 01:52 PM

if anyone can do it once more i got the last one wrong lol i noticed the edit after i answered it

x/2-y/3=5/6
x/5-y/4=-71/10

**yeKciM** · August 12th 2010, 02:09 PM

Originally Posted by mafai44

if anyone can do it once more i got the last one wrong lol i noticed the edit after i answered it

x/2-y/3=5/6
x/5-y/4=-71/10

here's using substitution :

$\displaystyle \frac {x}{2}- \frac {y}{3} = \frac {5}{6}$
$\displaystyle \frac {x}{5}- \frac {y}{4} = -\frac {13}{10}$

first one multiply with (6) and second with (20)

and you will get :

$3x-2y=5$
$4x-5y=-26$

now from any ( i'll do first ) you can express either "x" or "y" (i'll do "x" )

$\displaystyle 3x-2y=5 \Rightarrow 3x=2y+5 \Rightarrow x= \frac {2y+5}{3}$

now that "x we put in second equation ...

$\displaystyle 4(\frac {2y+5}{3} ) -5y = -26$

now multiply with 3 :

$\displaystyle 4(2y+5) -15y = -78 \Rightarrow 8y+20-15y = -78 \Rightarrow 7y = 98 \Rightarrow y = 14$

now that "y" we will put in first equation

$3x-28 = 5 \Rightarrow 3x = 33 \Rightarrow x=11$

P.S. what ? now second equation is with -71/10 , where did go -13/10 ?

Thread: solve by substitution

LinkBack

Thread Tools

Display