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Can someone help me with Algebra 2?

Can Someone Help Me With Algebra 2 class=

Sagot :

anuksha0456

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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