Question

linear systems

1. translate each sentence into an equation. tell how you are assigning the two variables.

a) the perimeter of a basketball court is 40 m.
b) the average of two numbers is 16.
c) the value of the quarters and loonies in a vending machine is $37.
d) the total receipts from adult tickets at $20 each and student tickets at $12 each was $9250.

1. use graphing to find the point of intersection of each pair of lines.

a) \\(x - y = 4\\)
\\(3x + 2y = 7\\)
b) \\(y = \frac{1}{2}x - 4\\)
\\(y = 2x - 1\\)
c) \\(x + y - 4 = 0\\)
\\(5x - y - 8 = 0\\)

1. solve each linear system using the method of substitution.

a) \\(x + 4y = 6\\)
\\(2x - 3y = 1\\)
b) \\(y = 6 - 3x\\)
\\(y = 2x + 1\\)
c) \\(5x - y = 4\\)
\\(3x + y = 4\\)

1. solve by elimination. check each solution.

a) \\(x + y = 55\\)
\\(2x - y = -4\\)
b) \\(2a + b = 5\\)
\\(a - 2b = 10\\)
c) \\(4k + 3h = 12\\)
\\(4k - h = 4\\)
d) \\(5a - 2b = 5\\)
\\(3a + 2b = 19\\)

1. explain why the following linear system has no solution.

\\(y - 2x = 1\\)
\\(y = 2x + 3\\)

1. use substitution or elimination to find the point of intersection of each pair of lines. round your answers to the nearest tenth.

a) \\(y = x - 5\\)
\\(x + 2y = 10\\)
b) \\(2x + 5y + 20 = 0\\)
\\(5x - 3y + 15 = 0\\)
c) \\(y = 7x\\)
\\(3y = 5x - 2\\)

1. find the values of \\(a\\) and \\(b\\) in the diagram shown.

1. a boat took 5 h to travel 60 km up a river, against the current. the return trip took 3 h. find the speed of the boat in still water and the speed of the current.

1. what volumes of 60% hydrochloric acid solution and 30% hydrochloric acid solution must be mixed to make 125 ml of 36% hydrochloric acid solution?

1. solve the linear system.

\\(\frac{x - 2}{3} + \frac{y + 1}{5} = 2\\)
\\(\frac{x + 2}{7} - \frac{y + 5}{3} = -2\\)

Explanation:

Identify the selected questions

The image contains a worksheet on linear systems with several circled questions:

• Question 2b: Solve by graphing.
• Question 3a, 3c: Solve by substitution.
• Question 4b, 4d: Solve by elimination.
• Question 7: Find the values of \(a\) and \(b\) in the diagram.
• Question 9: Mixture word problem.

We will solve these specific circled questions step-by-step.

Solve Question 2b

We need to find the intersection of:

1. \(y = \frac{1}{2}x - 4\)
2. \(y = 2x - 1\)

Equating the two expressions for \(y\):

$$\frac{1}{2}x - 4 = 2x - 1$$

Multiply by 2:

$$x - 8 = 4x - 2$$

$$-6 = 3x \implies x = -2$$

Substitute \(x = -2\) back into equation 2:

$$y = 2(-2) - 1 = -5$$

The intersection point is \((-2, -5)\).

Solve Question 3a and 3c

For 3a:

1. \(x + 4y = 6 \implies x = 6 - 4y\)
2. \(2x - 3y = 1\)

Substitute \(x\) into equation 2:

$$2(6 - 4y) - 3y = 1$$

$$12 - 8y - 3y = 1 \implies 12 - 11y = 1 \implies 11y = 11 \implies y = 1$$

Substitute \(y = 1\) into \(x = 6 - 4y\):

$$x = 6 - 4(1) = 2$$

Solution: \((2, 1)\).

For 3c:

1. \(5x - y = 4 \implies y = 5x - 4\)
2. \(3x + y = 4\)

Substitute \(y\) into equation 2:

$$3x + (5x - 4) = 4$$

$$8x - 4 = 4 \implies 8x = 8 \implies x = 1$$

Substitute \(x = 1\) into \(y = 5x - 4\):

$$y = 5(1) - 4 = 1$$

Solution: \((1, 1)\).

Solve Question 4b and 4d

For 4b:

1. \(2a + b = 5\)
2. \(a - 2b = 10\)

Multiply equation 1 by 2:

$$4a + 2b = 10$$

Add to equation 2:

$$(4a + 2b) + (a - 2b) = 10 + 10 \implies 5a = 20 \implies a = 4$$

Substitute \(a = 4\) into equation 1:

$$2(4) + b = 5 \implies 8 + b = 5 \implies b = -3$$

Solution: \((a, b) = (4, -3)\).

For 4d:

1. \(5a - 2b = 5\)
2. \(3a + 2b = 19\)

Add the two equations directly:

$$(5a - 2b) + (3a + 2b) = 5 + 19 \implies 8a = 24 \implies a = 3$$

Substitute \(a = 3\) into equation 2:

$$3(3) + 2b = 19 \implies 9 + 2b = 19 \implies 2b = 10 \implies b = 5$$

Solution: \((a, b) = (3, 5)\).

Solve Question 7

From the geometry diagram:

• The angle \(124^\circ\) and the angle \((3a - 2b)^\circ\) form a linear pair (straight line):

$$124 + (3a - 2b) = 180 \implies 3a - 2b = 56$$

• The angle \((2a + 3b)^\circ\) and the angle \(124^\circ\) are vertically opposite angles:

$$2a + 3b = 124$$

We have the system:

1. \(3a - 2b = 56\)
2. \(2a + 3b = 124\)

Multiply equation 1 by 3 and equation 2 by 2:

$$9a - 6b = 168$$

$$4a + 6b = 248$$

Add them:

$$13a = 416 \implies a = 32$$

Substitute \(a = 32\) into equation 2:

$$2(32) + 3b = 124 \implies 64 + 3b = 124 \implies 3b = 60 \implies b = 20$$

Solution: \(a = 32\), \(b = 20\).

Solve Question 9

Let \(x\) be the volume of \(60\%\) solution and \(y\) be the volume of \(30\%\) solution.

1. Total volume:

$$x + y = 125 \implies y = 125 - x$$

1. Pure acid content:

$$0.60x + 0.30y = 0.36(125)$$

$$0.60x + 0.30(125 - x) = 45$$

$$0.60x + 37.5 - 0.30x = 45$$

$$0.30x = 7.5 \implies x = 25\text{ mL}$$

$$y = 125 - 25 = 100\text{ mL}$$

Solution: \(25\text{ mL}\) of \(60\%\) solution and \(100\text{ mL}\) of \(30\%\) solution.

Answer:

Question 2b

The point of intersection is:

$$(-2, -5)$$

Question 3

a) The solution is:

$$(x, y) = (2, 1)$$

c) The solution is:

$$(x, y) = (1, 1)$$

Question 4

b) The solution is:

$$(a, b) = (4, -3)$$

d) The solution is:

$$(a, b) = (3, 5)$$

Question 7

The values are:

$$a = 32,\quad b = 20$$

Question 9

The required volumes are:

• \(25\text{ mL}\) of the \(60\%\) hydrochloric acid solution.
• \(100\text{ mL}\) of the \(30\%\) hydrochloric acid solution.