Question

equations by substitution
solve by substitution.

1. $x + 3y = 11$

$2x - 3y = 13$

1. $5x - y = 16$

$5x + 2y = 13$

1. $3x - 3y = 9$

$2x - 4y = -16$

1. $-x + 3y = -11$

$x + 7y = 16$

1. $-2x + 3y = -15$

$5x - 10y = 5$

1. $y = -3x + 21$

$2x - 7y = -9$

1. $x = 4y$

$4x + 2y = -36$

1. $14x - y = -12$

$5x - 5y = 5$

1. $-3x - y = 10$

$3x + 6y = 0$

1. $y = 7x - 1$

$40x - 7y = 79$

1. $x = -6y$

$y = \frac{2}{3}x + 5$

1. $9x + 5y = 7$

$y = \frac{2}{5}x - 3$

1. $5x + 2y = -1$

$-4x - y = -4$

1. $2x + \frac{y}{4} = -8$

$y = 8x$
write a system of equations to model each situation. solve the system by substitution. check the solution.

1. todd purchased two cds and three cassette tapes for $75. paulena went to the same music store and purchased one cd and five cassette tapes for $76. if all cds are the same price and all the cassette tapes are the same price, find the cost of each cd and each cassette tape.

each cd costs: $21, each cassette tape cost: $11.

1. one service station charges $35 per hour for labor and a towing fee of $30. another service station charges $45 per hour for labor without a towing fee. find the amount of time for which both companies charge the same amount.

Explanation:

Let's solve problem 1 as an example:

Step1: Add the two equations to eliminate \( y \)

We have the system:

$$
LATEXBLOCK0
$$

Adding the two equations: \((x + 3y) + (2x - 3y) = 11 + 13\)
Simplify: \(x + 2x + 3y - 3y = 24\) which gives \(3x = 24\)

Step2: Solve for \( x \)

Divide both sides of \( 3x = 24 \) by 3: \(x = \frac{24}{3} = 8\)

Step3: Substitute \( x = 8 \) into the first equation to find \( y \)

Substitute \( x = 8 \) into \( x + 3y = 11 \): \(8 + 3y = 11\)
Subtract 8 from both sides: \(3y = 11 - 8 = 3\)
Divide by 3: \(y = \frac{3}{3} = 1\)

Step1: Define variables

Let \( t \) be the time in hours and \( C \) be the total cost.

For the first service station: \( C = 35t + 30 \)

For the second service station: \( C = 45t \)

Step2: Set the equations equal to each other

Since we want to find when the costs are the same, set \( 35t + 30 = 45t \)

Step3: Solve for \( t \)

Subtract \( 35t \) from both sides: \( 30 = 45t - 35t \)
Simplify: \( 30 = 10t \)
Divide by 10: \( t = \frac{30}{10} = 3 \)

Answer:

The solution is \( x = 8 \), \( y = 1 \)

For problem 16: