2x + 5y = -3 ⇒ 2x + 5y = -3
1x + 8y = 4 ⇒ <u>2x + 16y = 8
</u> -<u>11y</u> = <u>-11 </u>
-11 -11
y = 1
2x + 5(1) = -3
2x + 5 = -3
<u> -5 -5</u>
<u>2x</u> = <u>-8</u>
2 2
x = -4
(x, y) = (-4, 1)
2x + 1y = 7 ⇒ 2x + 1y = 7
1x - 2y = -14 ⇒ <u>2x - 4y = -28</u>
<u>5y</u> = <u>35</u>
5 5
y = 7
2x + 7 = 7
<u> -7 -7</u>
<u>2x</u> = <u>0</u>
2 2
x = 0
(x, y) = (0, 7)
1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:


Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹
If you would like to solve p = r - c for c, you can do this using the following steps:
p = r - c /+c
p + c = r - c + c
p + c = r /-p
p + c - p = r - p
c = r - p
The correct result would be c = r - p.
Infinitely many solutions, you need to do the distributive property.