{"id":51,"date":"2014-10-21T03:50:59","date_gmt":"2014-10-21T03:50:59","guid":{"rendered":"https:\/\/blue-mathbelt.marjoriesayer.com\/?page_id=51"},"modified":"2017-07-23T09:07:49","modified_gmt":"2017-07-23T09:07:49","slug":"week-2-equations-answers","status":"publish","type":"page","link":"https:\/\/blue-mathbelt.marjoriesayer.com\/?page_id=51","title":{"rendered":"Week 2: Equations – Answers"},"content":{"rendered":"

Week 2: Equations – Day 5<\/strong><\/p>\n

Deriving Equations<\/p>\n

When deriving equations, there are three steps:
\n– assign variables to quantities
\n– figure out what principle, law or rule is involved
\n– use the principle, law or rule to set up an equation involving the variables<\/p>\n

1. Set up the equation for solving for the intersection between the line joining (0, 0) and (1, 4) and the line x + y = 1.<\/p>\n

(In this problem the variables are already set as x and y. The principle involved is that any two lines that are not parallel intersect at one point. The two equations for the lines will agree at that point.)<\/p>\n

The line joining (0, 0) and (1, 4) has slope (4-0)\/(1-0) = 4 and y-intercept 0. So its equation is y = 4x. To find the intersection between the two lines we can either substitute y = 4x into x + y = 1 to get:<\/p>\n

x + 4x = 1<\/p>\n

Or we can solve both equations for y and set them equal to one another:<\/p>\n

x + y = 1 corresponds to y = – x + 1<\/p>\n

so the equation for the intersection is:<\/p>\n

– x + 1 = 4x<\/p>\n

2. Set up the equation for solving for the zeros of the function f(x) = 2x3<\/sup> + 4x2<\/sup> – 17x – 3.<\/p>\n

2x3<\/sup> + 4x2<\/sup> – 17x – 3 = 0<\/p>\n

3. Set up the equation for solving for the time it takes to drive 432 kilometers at a speed of 65 kilometers per hour.<\/p>\n

Let T = time in hours.
\nLet D = distance = 432 km.
\nLet S = speed = 65 km\/h. <\/p>\n

When moving at constant speed, Distance = Speed x Time.
\nTherefore, T = D\/S<\/p>\n

Let T = time in hours. Then:
\nT = 432\/65. <\/p>\n

4. An isosceles triangle has two angles that are the same: 15 degrees. Set up the equation for solving for the other angle of the triangle.<\/p>\n

Let the unknown angle be X.
\nPrinciple: the sum of angles of a triangle is 180 degrees. <\/p>\n

Equation: X + 15 + 15 = 180<\/p>\n

5. Set up the equation for solving for the principal amount of a loan, if the amount of interest owed in one year is $700 and the interest rate is 5% per year.<\/p>\n

Let the principal amount = P.
\nLet the interest = I = 700.
\nLet the rate = R = 0.05
\nLet the time period = T = 1 year. <\/p>\n

The principle or rule involved in this problem is that: Interest = Principal x Rate x Time<\/p>\n

Therefore I = PRT<\/p>\n

And P = I\/RT = 700\/0.05\/1. <\/p>\n

\n

 <\/p>\n

Week 2: Equations – Day 4<\/strong><\/p>\n

Without solving these equations, predict the maximum and minimum number of solutions:<\/p>\n

1. 3x + 54 = 8x – 6<\/p>\n

This equation describes the intersection of two lines, y = 3x + 54 and y = 8x – 6. The lines have different slopes (3 and 8), are therefore not parallel, and will intersect at one point. There is exactly one solution to this equation.<\/p>\n

2. 4x – 7 = 4x + 20<\/p>\n

This equation describes the intersection of y = 4x – 7 and y = 4x + 20. These two lines have the same slope, 4, and are therefore parallel. They are not the same line, because they have different y-intercepts. Therefore there are exactly zero solutions (no solutions) to this equation.<\/p>\n

3. 5 + 4x = 3x2<\/sup> + 6x<\/p>\n

This equation can be rearranged to be the zero of a polynomial of degree 2:<\/p>\n

0 = 3x2<\/sup> + 6x – 4x – 5 = 3x2<\/sup> + 2x – 5<\/p>\n

A polynomial of degree 2 has at most 2 real zeros. There might be 0, 1, or 2 solutions.<\/p>\n

4. 10x5<\/sup> – 47x4<\/sup> + 3x3<\/sup> – x2<\/sup> + 8 = 0<\/p>\n

This equation is the zero of a polynomial of degree 5. Because this polynomial has odd degree, there must be at least one solution. There might be 1, 2, 3, 4, or 5 solutions.<\/p>\n

5. 1 = x2<\/sup> + 3<\/p>\n

This equation is a zero of a polynomial of degree 2:<\/p>\n

0 = x2<\/sup> + 2<\/p>\n

The above equation has no real solutions.<\/p>\n

\n

 <\/p>\n

Week 2: Equations – Day 3<\/strong><\/p>\n

Equations Involving Exponential Expressions<\/p>\n

Exponential expressions have two basic parts: exponent and base. The simplest equations involve one or the other.<\/p>\n

1. If 2300<\/sup> x 2n<\/sup> = 2501<\/sup> what is n?<\/p>\n

The left hand side is equivalent to:
\n2300+n<\/sup>
\nNow the bases are the same on both sides, so we can equate exponents according to the rules of exponents:
\n300 + n = 501<\/p>\n

n = 201.<\/p>\n

2. If 2100<\/sup> x 3100<\/sup> = a100<\/sup> what is a?<\/p>\n

The left hand side is the product of two bases with the same exponent. So it combines to:
\n(2 x 3)100<\/sup>
\nTherefore a = 6.<\/p>\n

3. If 100n<\/sup> x 53<\/sup> = 20n<\/sup> x 563<\/sup> what is n?<\/p>\n

100 = 22<\/sup> x 52<\/sup>
\nTherefore the left hand side becomes:
\n22n<\/sup> x 52n + 3<\/sup>
\n20 = 22<\/sup> x 5
\nTherefore the right hand side becomes:
\n22n<\/sup> x 5n + 63<\/sup>
\nThe factors involving 2 are the same on both sides, and cancel out.
\nWe are left with the 5s. We can equate the exponents of 5:
\n2n + 3 = n + 63<\/p>\n

Therefore, n = 60.
\nThere are many ways to do this problem.<\/p>\n

4. If 249<\/sup> x 2n<\/sup> = 1\/4, what is n?<\/p>\n

To solve the equation, we have to group terms with the same base.\u0010
\nThe left hand side becomes 249+n<\/sup> and the right hand becomes to 2-2<\/sup>.
\nEquating exponents:
\n49 + n = -2<\/p>\n

n = -51.<\/p>\n

5. If (2n<\/sup>)10<\/sup> = 326<\/sup> what is n?<\/p>\n

The left hand side is 210n<\/sup>.
\nThe right hand side is (25<\/sup>)6<\/sup> = 230<\/sup><\/p>\n

Equating exponents, we have:
\n10n = 30<\/p>\n

n = 3.<\/p>\n

\n

 <\/p>\n

Week 2: Equations – Day 2<\/strong><\/p>\n

Review of easy equation-solving techniques.<\/p>\n

1. How many solutions are there to:<\/p>\n

1\/(x2<\/sup> + 1)<\/sub> = 0<\/p>\n

There are no solutions. The only way there is a solution is if the numerator is zero. The numerator is 1 which is never zero, so there are no solutions.\u00a0The denominator is never zero, so you don’t have to worry about undefined values of x.<\/p>\n

2. What are the solutions of:<\/p>\n

(x – 2)\/(x2<\/sup> + 1)<\/sub> = 0<\/p>\n

The only solution is if (x – 2) = 0, or x = 2. The denominator causes no problems because it is always positive.<\/p>\n

3. What are the solutions of:<\/p>\n

x2<\/sup> – 28 = 3x<\/p>\n

There are two standard ways to solve a problem like this.<\/p>\n

One: turn it into an equation equal to zero and factor: x2<\/sup> – 3x – 28 =0
\nwhich factors into (x – 7)(x + 4) = 0 and there are two solutions, x = 7 and x = -4.<\/p>\n

Two: complete the square. This method works even if you can’t factor. If there are no solutions, this method will expose that fact.<\/p>\n

Bring the x-terms to one side and the constant to the other:
\nx2<\/sup> – 3x = 28
\nComplete the square on the left hand side:
\nx2<\/sup> – 3x + 9\/4 – 9\/4 = 28
\n(x – 3\/2)2<\/sup> = 28 + 9\/4 = (112 + 9)\/4 = 121\/4
\nTaking square roots of both sides:
\nx – 3\/2 = + or – 11\/2, or x = 3\/2 + or – 11\/2 = 14\/2 or -8\/2 = 7 or -4.<\/p>\n

4. What are the solutions of:<\/p>\n

(x + 3)2<\/sup> = 25<\/p>\n

Like method Two above, take square roots of both sides:
\nx + 3 = + or – 5 or x = – 3 + or – 5 which is x = 2 or x = -8.
\nIt’s important to check both values of x in the original equation. They both work.<\/p>\n

5. Solve:<\/p>\n

\"Screenshot_10_21_14__2_21_PM\"<\/a><\/p>\n

Cube both sides of the equation to get:<\/p>\n

x + 4 = 27
\nx = 23
\nPlugging into the original equation works, and so this is a valid solution.<\/p>\n

\n

 <\/p>\n

Week 2: Equations – Day 1<\/p>\n

1. \u00a0 3x – 3 = 5 – x<\/p>\n

Adding x to both sides: 4x – 3 = 5
\nAdding 3 to both sides: 4x = 8
\nx = 2<\/p>\n

2. \u00a0 1\/2 – 1\/x = 1\/x
\nNote: x=0 cannot be a solution.
\nMultiplying both sides by 2x: x – 2 = 2
\nx = 4<\/p>\n

3. x4<\/sup> – 1 = 0<\/p>\n

Factoring: (x2<\/sup> + 1)(x2<\/sup> – 1)
\nAnd again: (x2<\/sup> + 1)(x + 1)(x – 1) = 0
\nSolutions are x = -1 and x = 1<\/p>\n

4. 5x2<\/sup> – 3 = 2x<\/p>\n

Subtracting 3x from both sides: 5x2<\/sup> – 2x – 3 = 0<\/p>\n

Multiply coefficients of x2<\/sup> and 1 gives 5*(-3) = -15. Factors of -15 that add up to -2 are -5 and 3.
\n5x2<\/sup> -5x + 3x – 3 = 0
\n5x(x – 1) + 3(x – 1) = (5x + 3)(x – 1) = 0
\nSolutions are x = -3\/5 and x = 1.<\/p>\n

5. (x2<\/sup> – 2x – 3)\/(x + 1) = 0<\/p>\n

Note: x = -1 cannot be a solution.
\nMultiply both sides by (x + 1): x2<\/sup> – 2x -3 = 0
\nFactoring: Factors of -3 that add up to -2 are -3 and 1: (x – 3)(x + 1) = 0
\nSolution is only x = 3.
\nx = -1 cannot be a solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Week 2: Equations – Day 5 Deriving Equations When deriving equations, there are three steps: – assign variables to quantities – figure out what principle, law or rule is involved – use the principle, law or rule to set up an equation involving the variables 1. Set up the equation for solving for the intersection […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":10,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-51","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/pages\/51","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=51"}],"version-history":[{"count":10,"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/pages\/51\/revisions"}],"predecessor-version":[{"id":88,"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/pages\/51\/revisions\/88"}],"up":[{"embeddable":true,"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/pages\/10"}],"wp:attachment":[{"href":"https:\/\/blue-mathbelt.marjoriesayer.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=51"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}