Please help with:
$\displaystyle X + 7Y = 6 $
$\displaystyle 3X + 5Y = 14$
I still cant get my head around this!
Thanks in advance!
ADY
For example multiply the first equation (both sides) by 3: $\displaystyle 3X + 21Y = 18 $ and subtract from this the other equation: $\displaystyle (3-3)X+(21-5)Y=18-14 \Leftrightarrow 16Y=4\Leftrightarrow Y=\frac {4}{16}=\boxed{\frac 14=0.25}$ which you can substitute in one of the original equation: $\displaystyle X+7\cdot\frac 14=6\Leftrightarrow X=6-\frac 74=\boxed{\frac {17}4=4.25}$.
try this, substitution method
$\displaystyle X + 7Y = 6 $
$\displaystyle 3X + 5Y = 14$
$\displaystyle x=6-7y$
$\displaystyle 3(6-7y)+5y=14$
$\displaystyle 18-21y+5y=14$
$\displaystyle 18-14=21y-5y$
$\displaystyle 4=16y$
$\displaystyle y=\frac{1}{4}$
then substitute
$\displaystyle x+7y=6$
$\displaystyle x+\frac{7}{4}=6$
$\displaystyle x=6-\frac{7}{4}$
$\displaystyle x=\frac{24-7}{4}$
$\displaystyle x=\frac{17}{4}$
check
$\displaystyle \frac{17}{4}+7{1}{4}=6$
$\displaystyle \frac{24}{4}=6$
$\displaystyle 6=6$
$\displaystyle 3x+5y=14$
$\displaystyle 3*\frac{17}{4}+5*\frac{1}{4}=14$
$\displaystyle \frac{51+5}{4}=14$
$\displaystyle \frac{56}{4}=14$
$\displaystyle 56=56$