The break-even point is where the costs equal the revenues

Fixed Costs + Variable Costs = Revenues

1500 + 50x = 75x

Solving for x results in x = 60 boats sold each month to break even.

The commission she gets for selling seven cars is $500 * 7 = $3,500 and added to the salary of $1,000 yields $4,500 for the month.

Add the two equations to get 3x = 9, so x = 3.  Substitute the value of x into one of the equations to find the value of y; therefore x = 3 and y = –2, so their sum is 1.

Substitute the values of x and y into the given expression:

2(1/3) + 3(1/2)

= 2/3 + 3/2

= 4/6 + 9/6

= 13/6

The expression is undefined for exactly those values of \(\displaystyle x\) which yield a denominator of 0 - that is, for which 

\(\displaystyle x^{2} +9 =0\)

However, for all real \(\displaystyle x\), 

\(\displaystyle x^{2} \geq 0\) ,

and, subsequently,

\(\displaystyle x^{2} +9 \geq 9 >0\) 

meaning the denominator is always positive. Therefore, the expression is defined for all real values of \(\displaystyle x\).

Let \(\displaystyle x\) be the number of ten-dollar bills Albert has; then he has \(\displaystyle 13-x\) five-dollar bills.

He then has \(\displaystyle 10x + 5(13-x)\) dollars in his wallet, which must be greater than $100. Set up and solve an inequality:

\(\displaystyle 10x + 5(13-x) > 100\)

\(\displaystyle 10x + 65-5x > 100\)

\(\displaystyle 5x > 35\)

\(\displaystyle x>7\)

Therefore, the lowest whole number of ten-dollar bills that Albert can have is eight.

Expand both products, the left using distribution, the right using the binomial square pattern:

\(\displaystyle x (x+ 5) = \left ( x+2\right )^{2}\)

\(\displaystyle x \cdot x+x \cdot 5= x^{2} + 2 \cdot 2 \cdot x +2^{2}\right )\)

\(\displaystyle x^{2}+5 x = x^{2} + 4x +4\)

\(\displaystyle x^{2} +5 x-x^{2} = x^{2} + 4x +4 -x^{2}\)

\(\displaystyle 5 x= 4x +4\)

Note that the quadratic terms can be eliminated, yielding a linear equation.

\(\displaystyle 5 x-4x = 4x +4 -4x\)

\(\displaystyle x = 4\)

\(\displaystyle 4x + 2(x-7) = 2 (3x -5)\)

\(\displaystyle 4x + 2 \cdot x-2 \cdot7 = 2 \cdot 3x -2 \cdot5\)

\(\displaystyle 4x + 2 x-14 = 6x -10\)

\(\displaystyle \left (4 + 2 \right )x-14 = 6x -10\)

\(\displaystyle 6x-14 = 6x -10\)

\(\displaystyle 6x-6x-14 = 6x -6x -10\)

\(\displaystyle -14 = -10\)

This identically false statement alerts us to the fact that the original equation has no solution.

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

\(\displaystyle (2x-7)(x + 4) = 27\)

\(\displaystyle 2x \cdot x+ 2x \cdot 4 -7\cdot x -7\cdot 4 = 27\)

\(\displaystyle 2x^{2}+ 8x -7 x -28 = 27\)

\(\displaystyle 2x^{2}+ x -28 = 27\)

\(\displaystyle 2x^{2}+ x -28-27 = 27-27\)

\(\displaystyle 2x^{2}+ x- 55 = 0\)

Use the \(\displaystyle ac\) method to split the middle term into two terms; we want the coefficients to have a sum of 1 and a product of \(\displaystyle 2 (-55) = -110\). These numbers are \(\displaystyle -10,11\), so we do the following:

\(\displaystyle 2x^{2}-10x+ 11x- 55 = 0\)

\(\displaystyle 2x\left (x- 5 \right) + 11 \left ( x- 5 \right) = 0\)

\(\displaystyle \left (2x+ 11 \right) \left ( x- 5 \right) = 0\)

Set each expression equal to 0 and solve:

\(\displaystyle 2x + 11 = 0\)

\(\displaystyle 2x = -11\)

\(\displaystyle x = -5\frac{1}{2}\)

or 

\(\displaystyle x-5= 0\)

\(\displaystyle x = 5\)

The solution set is \(\displaystyle \left \{ -5 \frac{1}{2}, 5\right \}\).

First, rewrite the quadratic equation in standard form by distributing the \(\displaystyle x\) through the product on the left, then collecting all of the terms on the left side:

\(\displaystyle x\left (3x+7 \right ) =20\)

\(\displaystyle x\left \cdot \; 3x+x\left \cdot\; 7 =20\)

\(\displaystyle 3x ^{2}+7x =20\)

\(\displaystyle 3x ^{2}+7x -20=20 -20\)

\(\displaystyle 3x ^{2}+7x -20=0\)

Use the \(\displaystyle ac\) method to factor the quadratic expression \(\displaystyle 3x ^{2}+7x -20\); we are looking to split the linear term by finding two integers whose sum is 7 and whose product is \(\displaystyle 3 (-20) = -60\). These integers are \(\displaystyle -5,12\), so:

\(\displaystyle \left (3x ^{2}-5x \right ) + \left ( 12x -20 \right ) =0\)

\(\displaystyle x \left (3x-5 \right ) +4 \left ( 3x -5 \right ) =0\)

\(\displaystyle \left (x +4 \right )\left (3x-5 \right )=0\)

Set each expression equal to 0 and solve:

\(\displaystyle x+4 = 0\)

\(\displaystyle x=-4\)

or

\(\displaystyle 3x-5=0\)

\(\displaystyle 3x=5\)

\(\displaystyle x = \frac{5}{3} = 1\frac{2}{3}\)

The solution set is \(\displaystyle \left \{ -4, 1\frac{2}{3}\right \}\).