# Thread: [SOLVED] More Sim Equations

1. ## [SOLVED] More Sim Equations

$X + 7Y = 6$
$3X + 5Y = 14$

I still cant get my head around this!

2. For example multiply the first equation (both sides) by 3: $3X + 21Y = 18$ and subtract from this the other equation: $(3-3)X+(21-5)Y=18-14 \Leftrightarrow 16Y=4\Leftrightarrow Y=\frac {4}{16}=\boxed{\frac 14=0.25}$ which you can substitute in one of the original equation: $X+7\cdot\frac 14=6\Leftrightarrow X=6-\frac 74=\boxed{\frac {17}4=4.25}$.

$X + 7Y = 6$
$3X + 5Y = 14$

I still cant get my head around this!

$3x+21y=18$ -------------1

$3x+5y=14$ --------------2

2-1

Therefore , 16y=4 , y=1/4 , then continue to solve for x .

4. $
x = \frac{17}{4}, y = \frac{1}{4}
$

?

$
x = \frac{17}{4}, y = \frac{1}{4}
$

?
Yeah.

$X + 7Y = 6$
$3X + 5Y = 14$

I still cant get my head around this!

try this, substitution method

$X + 7Y = 6$

$3X + 5Y = 14$

$x=6-7y$

$3(6-7y)+5y=14$

$18-21y+5y=14$

$18-14=21y-5y$

$4=16y$

$y=\frac{1}{4}$

then substitute

$x+7y=6$

$x+\frac{7}{4}=6$

$x=6-\frac{7}{4}$

$x=\frac{24-7}{4}$

$x=\frac{17}{4}$

check

$\frac{17}{4}+7{1}{4}=6$

$\frac{24}{4}=6$

$6=6$

$3x+5y=14$

$3*\frac{17}{4}+5*\frac{1}{4}=14$

$\frac{51+5}{4}=14$

$\frac{56}{4}=14$

$56=56$