1. Solving linear equations:
(a) x = 4.
(b) r = 2/3.
2. Proportion:
(a) When x = 40, y = 100.
(b) When x = 40, y = 4.
3. System of equations:
(a) r = 8, s = 3.
(b) p = 5, q = -2.
4. Graphing lines:
(a) The slope of the line through (3, 4) and (-1, 3) is 1/4.
(b) The slope of the line 3x - 4y = 7 is 3/4.
(c) The slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.
5. Introductory quadratics:
(a) The solutions to the equation 4x² = 81 are -9/2, and 9/2.
(b) The solutions to the equation x² + 8x + 12 are -6, and -2.
(c) The solutions to the equation x² - 3x - 88 are -8, and 11.
6. The weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.
7. x = 3.
1. Solving linear equations:
(a) 3x - 7 = 9 - x,
or, 3x + x = 9 + 7,
or, 4x = 16,
or, x = 16/4 = 4.
Thus, x = 4.
(b) (7 - 2r)/3 = 4r,
or, 7 - 2r = 3*4r = 12r,
or, -2r - 12r = -7,
or, -14r = -7,
or, r = -7/-14 = 1/2.
Thus, r = 1/2.
2. Proportion:
(a) x and y directly proportion, means x/y = constant.
When x = 8, y = 20.
Thus, x/y = constant, or, 8/20 = constant, or, constant = 0.4.
When x = 40,
x/y = 0.4,
or, y = x/0.4 = 40/0.4 = 100.
Thus, when x = 40, y = 100.
(b) x and y indirectly proportion, means xy = constant.
When x = 8, y = 20.
Thus, xy = constant, or, 8*20 = constant, or, constant = 160.
When x = 40,
xy = 160,
or, 40y = 160,
or, y = 160/40 = 4.
Thus, when x = 40, y = 4.
3. System of equations:
(a) r - s = 5 ...(i)
3r - 5s = 9 ... (ii)
3*(i) - (ii) gives:
3r - 3s = 15
3r - 5s = 9
(-) (+) (-)
_________
2s = 6,
or, s = 3.
Substituting in (i), we get
r - s = 5,
or, r - 3 = 5,
or, r = 8.
Thus, r = 8, s = 3.
(b) 3p + 7q = 1 ...(i).
5p = 14q + 53,
or, 5p - 14q = 53 ...(ii).
2*(i) + (ii) gives:
6p + 14q = 2
5p - 14q = 53
____________
11p = 55,
or, p = 5.
Substituting in (i), we get:
3p + 7q = 1,
or, 3*5 + 7q = 1,
or, 7q = 1 - 15 = -14,
or, q = -14/7 = -2.
Thus, p = 5, q = -2.
4. Graphing lines:
(a) Slope of the line through (3, 4) and (-1, 3) is,
m = (4 - 3)/(3 - (-1)),
or, m = 1/4.
Thus, the slope of the line through (3, 4) and (-1, 3) is 1/4.
(b) The graph given: 3x - 4y = 7.
Representing in the slope-intercept form, y = mx + b, gives:
3x - 4y = 7,
or, 4y = 3x - 7,
or, y = (3/4)x + (-7/4).
Thus, the slope of the line 3x - 4y = 7 is 3/4.
(c) Slope of the line through (5, -2) and (-1, 6) is,
m = (6 - (-2))/(-1 - 5),
or, m = 8/(-6) = -4/3.
Substituting m = -4/3 in the slope-intercept form, y = mx + b, gives:
y = (-4/3)x + b.
Substituting y = 6, and x = -1 gives:
6 = (-4/3)(-1) + b,
or, b = 6 - 4/3 = 14/3.
Thus, the slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.
5. Introductory quadratics:
(a) 4x² = 81,
or, 4x² - 81 = 0,
or (2x)² - 9² = 0,
or, (2x + 9)(2x - 9) = 0.
Either, 2x + 9 = 0 ⇒ x = -9/2,
or, 2x - 9 = 0 ⇒ x = 9/2.
Thus, the solutions to the equation 4x² = 81 are -9/2, and 9/2.
(b) x² + 8x + 12 = 0,
or, x² + 2x + 6x + 12 = 0,
or, x(x + 2) + 6(x + 2) = 0,
or, (x + 6)(x + 2) = 0.
Either, x + 6 = 0 ⇒ x = -6,
or, x + 2 = 0 ⇒ x = -2.
Thus, the solutions to the equation x² + 8x + 12 are -6, and -2.
(c) x² - 3x - 88 = 0,
or, x² - 11x + 8x - 88 = 0,
or, x(x - 11) + 8(x - 11) = 0,
or, (x + 8)(x - 11) = 0.
Either, x + 8 = 0 ⇒ x = -8,
or, x - 11 = 0 ⇒ x = 11.
Thus, the solutions to the equation x² - 3x - 88 are -8, and 11.
6. We assume the weight of one orange, one apple, and one banana to be x, y, and z units respectively.
Thus, we have:
3x + 2y + z = 15 ... (i)
5x + 7y + 2z = 44 ... (ii)
x + 3y + 5z = 26 ... (iii)
2(i) - (ii) gives:
6x + 4y + 2z = 30
5x + 7y + 2z = 44
(-) (-) (-) (-)
______________
x - 3y = -14 ... (iv)
5(i) - (iii) gives:
15x + 10y + 5z = 75
x + 3y + 5z = 26
(-) (-) (-) (-)
________________
14x + 7y = 49 ... (v)
14(iv) - (v) gives:
14x - 42y = -196
14x + 7y = 49
(-) (-) (-)
_____________
-49y = -245,
or, y = 5.
Substituting in (v), we get:
14x + 7y = 49,
or, 14x + 35 = 49,
or, x = 14/14 = 1.
Substituting x = 1 and y = 5 in (i), we get:
3x + 2y + z = 15,
or, 3 + 10 + z = 15,
or, z = 2.
Thus, the weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.
7. 3/(1 - (2/x)) = 3x,
or, 3/((x - 2)/x) = 3x,
or, 3x/(x - 2) = 3x,
or, 3x = 3x(x - 2),
or, 3x = 3x² - 6x,
or, 3x² - 9x = 0,
or, 3x(x - 3) = 0.
Either, 3x = 0 ⇒ x = 0,
or, x - 3 = 0 ⇒ x = 3.
Since we had a term 2/x, x cannot be 0.
Thus, x = 3.
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