Question 568
your first question has the same line, so we will solve your second. 

since you have two variables (two unknowns - x and y), they are giving you two equations so that you can solve for each variable by substitution.
<br>
step 1: write equations in linear format <b>y = mx + b</b>
so for <i>3x + 4y = 12</i>, we write it as <i>4y = -3x + 12</i>. 
for <i>6x + 14y = 30</i>, we write it as <i>14y = -6x + 30</i>
<br>
step 2: simplify your equations.
for the equation <i>4y = -3x + 12</i>, divide the whole equation by <b>4</b> so that the coefficient of y will equal 1. 
<i>4y/4 = -3x/4 + 12/4</i>
after dividing, this will be written as <i>y = -3x/4 + 3</i>
2nd equation: <i>14y = -6x + 30</i> will be divided by <b>14</b> and will be rewritten as <i>y = -3x/7 + 15/7</i>
<br>
step 3: substitute y in the first equation since there are two variables. 
since y = -3x/7 + 15/7, when we combine your 2nd equation with your 1st, we write it as <i>-3x/7 + 15/7 = -3x/4 + 3</i>.
<br>
step 4: bring x's to one side of the equation and your constants to one side.
<i>-3x/7 + 3x/4 = 3 - 15/7</i>
<br>
step 5: find the lowest common denominator to get rid of the fractions. thus, we will multiply the whole equation of -3x/7 + 3x/4 = 3 - 15/7 by <b>28</b> and get:
<i>-12x + 21x = 3(28) - 60
9x = 84 - 60
9x = 24
x = 24/9 = <b>8/3</b></i>
<br>
step 6: subsitute x into your original equation. we will use the 1st one <i>3x + 4y = 12
3(8/3) + 4y = 12
(the 3's cancel)
8 + 4y = 12
4y = 12 - 8
4y = 4
<b>y = 1</b></i>
<br>
Thus, <font color=red>x = 8/3 and y = 1</font>