The coupon will entitle him to one free croissant, so Jerry will pay for the large espresso, the large cappucino, and one butter croissant. The charge will be the sum of the three prices:

\(\displaystyle \begin{matrix} \$3.29 \\ \; \; 3.89\\ \underline{+2.29} \\ \$9.47 \end{matrix}\)

Use the distance formula to solve this problem.

\(\displaystyle A=(x_1,y_1), B=(x_2,y_2)\)

\(\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Plugging in our points A and B we get the following distance.

\(\displaystyle A = (3, 4) B = (2, 5)\)

\(\displaystyle d = \sqrt{(2 - 3)^{2} + (5 - 4)^{2}}\)

\(\displaystyle d = \sqrt{1 + 1} = \sqrt{2} = 1.41\)

First, take the first equation and solve for x in terms of y.

\(\displaystyle 2x + 4y = 10\)

\(\displaystyle -4y -4y\)

\(\displaystyle 2x = -4y + 10\)

\(\displaystyle x = -2y + 5\)

From here, substitute the equation you found for x into the second equation and solve for y. 

\(\displaystyle 5x - 3y = 11\)

\(\displaystyle 5 (-2y + 5) - 3y = 11\)

\(\displaystyle -10y + 25 - 3y = 11\)

\(\displaystyle -13y + 25 = 11\)

\(\displaystyle -25 -25\)

\(\displaystyle -13y = -14\)

\(\displaystyle y = 1.08\)

Now, substitute the y value found into the original equation and solve for x.

\(\displaystyle 2x + 4y = 10\)

\(\displaystyle 2x + 4(1.08) = 10\)

\(\displaystyle 2x + 4.32 = 10\)

\(\displaystyle -4.32 -4.32\)

\(\displaystyle 2x = 5.68\)

\(\displaystyle x = 2.85\)

To solve this system of equations by elimination you first need to multiply the first equation by -2. This will allow the y terms to cancel when equation 1 and equation 2 are adding together.

\(\displaystyle -2 \times \left \{ 5x - y = 7 \right \} = -10x + 2y = -14\)

\(\displaystyle \left \{ 6x - 2y = 6 \right \} = 6x - 2y = 6\)

Now solve for x.

\(\displaystyle -4x = -8\)

\(\displaystyle x = 2\)

Substitute the x value into the orginial equation and solve for y.

\(\displaystyle 5x - y = 7\)

\(\displaystyle 5(2) - y = 7\)

\(\displaystyle 10 - y = 7\)

\(\displaystyle -10 -10\)

\(\displaystyle -y = -3\)

\(\displaystyle y = 3\) 

\(\displaystyle \left (\frac{3}{4}x + \frac{2}{5} \right ) - \left (\frac{1}{8}x - \frac{1}{10} \right )\)

\(\displaystyle =\frac{3}{4}x + \frac{2}{5} - \frac{1}{8}x + \frac{1}{10}\)

\(\displaystyle =\frac{3}{4}x- \frac{1}{8}x + \frac{2}{5} + \frac{1}{10}\)

\(\displaystyle =\frac{6}{8}x- \frac{1}{8}x + \frac{4}{10} + \frac{1}{10}\)

\(\displaystyle =\frac{5}{8}x + \frac{5}{10}\)

\(\displaystyle =\frac{5}{8}x + \frac{1}{2}\)

There is only one like-term in the expression given, which is \(\displaystyle a^2bc\).

\(\displaystyle -3a^2bc+3ab^2-4a^2b+2a^2bc\)

Add the like term.  The rest of the terms cannot be combined by adding or subtracting.

The correct answer is:

\(\displaystyle -a^2bc+3ab^2-4a^2b\)

Simplify first by eliminating the parentheses. To do this you will need to distribute the coefficient term outside the parentheses to each term within the parentheses. 

\(\displaystyle -7xy + 8y- 2x(y+3)+6(y-3)\)

Next, combine like terms.

\(\displaystyle =-7xy + 8y- 2xy-6x+6y-18\)

\(\displaystyle =-9xy + 14y-6x-18\)

Add the ones digits.

\(\displaystyle 9+2 = 11\)

Since this number is \(\displaystyle 10\) or greater, use the tens digit as a carry over.

Add the tens digits with the carry over.

\(\displaystyle 6+3+(1)=10\)

Add the hundreds digit with the carry over.  The hundreds digit for the second number in the problem is zero.

\(\displaystyle 3+0+(1)= 4\)

Since there is no carry over and there is no thousands digit for the second number, the thousands digit of the answer is just \(\displaystyle 2\).

Combine all the ones digits in chronological order from thousands to the ones digits.

The answer is:  \(\displaystyle 2401\)

In order to simplify this, we need to combine like-terms.  The coefficients of unlike terms cannot be combined.

Add \(\displaystyle 2a\) and \(\displaystyle 9a\).

\(\displaystyle 2a+9a = 11a\)

Add \(\displaystyle -ab^2\) and \(\displaystyle 4ab^2\).

\(\displaystyle -ab^2+4ab^2 = 3ab^2\)

Add \(\displaystyle 3ab\) and \(\displaystyle 6ab\).

\(\displaystyle 3ab+ 6ab =9ab\)

Combine all the terms.

The answer is:  \(\displaystyle 11a+3ab^2+9ab\)

In order to solve the problem, first evaluate \(\displaystyle 61-1 4\).

Borrow a one from the six in the tens digit in order to subtract the ones digit.

\(\displaystyle 11-4 = 7\)

The six in the tens digit becomes a five.  Subtract the five with the tens digit of the second number.

\(\displaystyle 5-1 =4\)

\(\displaystyle 61-1 4 = 47\)

Add \(\displaystyle 47 +23\).

Add the ones digit. 

\(\displaystyle 7+3 = 10\)

Since the number is ten or greater, use the tens digit as a carryover to the next calculation.

\(\displaystyle 4+2+(1) = 7\)

Combine the ones digits.

The answer is \(\displaystyle 70\).