Use substitution of x = 2 into the first equation to find y:

$$ \begin{align} 2(2) + 37 &= 7 \\ 4 + 3y &= 7 \\ 3y &= 3 \\ y &= 1 \end{align}$$

The point of intersection is (2, 1)

Solve the first equation for x, then substitute that into the second:

$$ \begin{align} x &= 2y + 4 \\ 3(2y + 4) + y &= 5 \\ 6y + 12 + y &= 5 \\ 7y &= -7 \\ y &= -1 \end{align}$$

Now use either equation to find x:

$$ \begin{align} x - 2y &= 4 \rightarrow x = 2y + 4 \\ x &= 2(-1) + 4 \\ x &= 2 \end{align}$$

The intersection point is (2, -7)

These equations are set up nicely for elimination of x through addition:

$$ \begin{matrix} 3x + 4y = 12 && \\ \underline{-3x - y = -4} && \leftarrow \text{ add}\\ \phantom{0000}3y = 8 && \\[5pt] \phantom{00000}y = \frac{8}{3} \end{matrix}$$

Now use either equation to find x:

$$ \begin{align} 3x + 4\left( \frac{8}{3} \right) &= 12 \\ 3x + \frac{32}{3} &= \frac{36}{3} \\ 3x &= \frac{4}{3} \\ x &= \frac{4}{9} \end{align}$$

The point of intersection is $\left( \frac{4}{9}, \frac{8}{3} \right)$

Subtracting these equations (either way) will eliminate y:

$$ \begin{matrix} x - y = 7 && \\ \underline{7x - y = 2} && \leftarrow \text{ subtract}\\ \phantom{0}-6x = 5 && \\ \phantom{00000}x = \frac{-5}{6} && \end{matrix}$$

Now use the first equation to get y:

$$ \begin{align} y &= x - 7 \\ y &= -\frac{5}{6} - \frac{42}{6} \\[3pt] y &= \frac{-47}{6} \end{align}$$

The point of intersection is $\left( \frac{-5}{6}, \frac{-47}{6} \right)$

We'll use elimination. Multiply the first equation by 2, then add them.

$$ \begin{matrix} -2x - 6y = 22 && \\ \underline{\phantom{0}2x + 5y = -8} && \leftarrow \text{ add}\\ \phantom{000000}y = 14 && \end{matrix}$$

Now use one of the other equations (either will work; choose the simplest one) to find x.

$$ \begin{align} -x - 3y &= 11 \\ x &= -3y - 11 \\ x &= -3(14) - 11 \\ x &= -42 - 11 = -53 \end{align}$$

The point of intersection is (-53, 14)

This system is perfectly set up to eliminate y using the transitive property:

$$ \begin{align} 3x - 11 &= 2x + 1 \\ x &= 12 \end{align}$$

Now use either equation to find y from x:

$$ \begin{align} y &= 3x - 11 \\ y &= 3(12) - 11 \\ y &= 36 - 11 = 25 \end{align}$$

The point of intersection is (12, 25)

Let's let the number of fives be f and the number of twenties be t. So we have   $f + t = 73.$ Then the number of dollars is 5 times the number of fives and 20 times the number of twenties, or   $5f + 20t = 725.$

We can solve this by substitution:

$$ \begin{align} f + t &= 73 \; \rightarrow \; f = 73 - t \\ 5(73 - t) + 20t &= 725 \\ 365 - 5t + 20t &= 725 \\ 15t &= 365 \\ t &= \frac{365}{15} = 24 \; \text{ twenties} \end{align}$$

Then the number of fives is

$$73 - 24 = 49 \; \text{ fives}$$

So the pile contains 49 fives and 24 twenties for a total of 73 bills summing to \$ 725.00

This one isn't exactly linear, but we can still do it. If you don't know what completing the square is, just skip this problem.

Our equations are (from the perimeter and area, respectively),

$$ \begin{align} 2x + 2y &= 320 \; \longrightarrow \; x + y = 60 \\[2pt] xy &= 5000 \end{align}$$

Substituting the former into the latter gives:

$$ \begin{align} x(160 - x) &= 5000 \\[2pt] -x^2 + 160 x &= 5000 \\[2pt] x^2 - 160 x &= -5000 \\[2pt] x^2 - 160 x + 80^2 &= 6400 - 5000 \leftarrow \text{compl. the square} \\[2pt] (x - 80)^2 &= 1400 \\[2pt] x &= 80 ± \sqrt{1400} \\[2pt] x &= 80 ± 37.42 \\[2pt] x &= 117.42, \; 42.58 \end{align}$$

These happen to add to 160, so they are our x and y. They yield a perimeter of 320 m and an area of 5000 m².

The dimensions of the field are 117.42 m × 42.58 m.

Let e and c be the number of pounds of expensive and cheap coffee, respectively. Our information consists of the total number of pounds (50) and the total number of dollars (\$5.50/lb. × 50 lbs). Our equations are

$$ \begin{align} e + c &= 50 \\[2pt] 9e + 4c &= 5.5(50) \; \leftarrow \; c = 50 - e \\[2pt] 9e + 4(50 - e) &= 275 \\[2pt] 9e + 200 - 4e &= 275 \\[2pt] 5e &= 75 \\[2pt] e &= 15 \; lbs. \\[2pt] c &= 35 \; lbs. \end{align}$$

The blender would mix 15 lb. of the expensive coffee with 35 lb. of the cheap to make 50 lb. of a blend that could sell for \$5.50 per pound.

Let's make w the number of milliliters (mL) of our weak (2%) solution and s the number of mL of the strong (25%) solution. The percent concentration (as a decimal) is the number of mL of glucose divided by the number of mL of the solution (250 mL).

We have two constraints or pieces of information. They are:

$$ \begin{align} w + s &= 250 \\[2pt] \frac{0.25 s + 0.02 w}{250} &= 0.04 \; \text{ or} \\[2pt] 0.25 s + 0.02 w &= 10 \end{align}$$

Now if we substitute w = 250 - s for w in the last equation, we can solve the system:

$$ \begin{align} 0.25 s + 0.02(250 - s) &= 10 \\[2pt] 0.25 s + 5 - 0.02 s &= 10 \\[2pt] 0.23 s &= 5 \\[2pt] s &= 21.74 \; mL \end{align}$$

then w = 250 - 21.74 = 228.26 mL

The solution is made by adding 21.74 mL of the strong (25%) solution to 228.26 mL of the weak (2%) solution.

In this question, we have two equations of the same type, both sums of costs of tickets of two types. Let s be the number of Redsox tickets and m be the number of Mariners tickets. Then we have

$$ \begin{align} 4s + 2m &= 500.00 \\ 3s + m &= 344.00 \end{align}$$

This system is best solved by elimination. We'll multiply the bottom equation by -2 and add:

$$ \begin{matrix} 4s + 2m = 500.00\phantom{0} && \\ \underline{3s + m = 344.00} && \leftarrow \times -2 \\ 4s + 2m = 500.00 && \\ \underline{-2(3s + m = 344)\phantom{000}} && \\ 4s + 2m = 500\phantom{00} && \\ \underline{-6s - 2m = -688\phantom{00}} && \leftarrow \text{ add} \\ \phantom{00}-2s = -188 && \\ \phantom{000000}s = \$ 94.00 && \end{matrix}$$

Then using either of the original equations, we get   $m = \$62.00.$

Redsox tickets: \$94.00, Mariners tickets: \$62.00

Let r be the number of dollars in the REIT and m be the number in the mutual fund. Then our two equations are:

$$ \begin{align} r + m &= 1000.00 \\[2pt] 0.055r + 0.075m &= 532 \end{align}$$

where 532.00 is the amount of interest earned by the end of the year (the amount over the initial \$10,000 investment. Now we can solve the system by substitution:

$$ \begin{align} r + m &= 1000.00 \rightarrow r = 1000 - m \\[2pt] 0.055(1000 - m) + 0.075m &= 532 \leftarrow \text{ substitute} \\[2pt] 55 - 0.055m + 0.075m &= 532 \\[2pt] 0.02m &= 477 \\[2pt] m &= \frac{477}{0.2} = \$2385.00 \\[2pt] r = 1000 - 2385 &= \$7615.00 \end{align}$$

The original investment was \$2,385.00 in the mutual find and \$7,615.00 in the REIT.

Let C be the speed of the car and T be the speed of the truck. We see that C = T + 5. Now we know that

$$s = \frac{d}{t} \longrightarrow t = \frac{d}{s},$$

where s = speed, d = distance and t = time. Because the times are identical, we can set d/s equal for both vehicles:

$$ \begin{align} \frac{300}{C} = \frac{280}{T} \\[4pt] \frac{300}{T + 5} = \frac{280}{T} \\[4pt] 300T = 280T + 1400 \\[2pt] 20T = 1400 \\[4pt] T = \frac{1400}{20} = 70 \; \text{mi./h} \\[2pt] C = 75 \; \text{mi./h} \end{align}$$

The speed of the car was 75 mi/h and the speed of the truck was 70 mi/h.

If we let n be the number of shirts, then the costs from company 1 and company 2 are:

$$ \begin{align} C_1 &= 8.5n + 40 \\[2pt] C_2 &= 9.25n + 24 \end{align}$$

Now we're interested in what n is when C₁ = C₂, so we set those equations equal to one another and solve:

$$ \begin{align} 8.5n + 40 &= 9.25n + 24 \\[2pt] 16 &= 0.75 \, n \\[2pt] n &= \frac{16}{0.75} = 21.33 \; \text{ shirts} \end{align}$$

So as long as 22 or more shirts are ordered, both companies are the same (in terms of cost, anyway). If fewer than 22 are ordered, company 2 is cheaper.