Most Efficient way to solve this problem?

<p>Sorry if this is the wrong forum but I was doing some math saw this problem and I was wondering if there was a faster way to solve it.</p>

<p>Case picture is unclear its 7 divided by (3x + 4) is equal to 7 divided by (6x-2)</p>

<p>7 _______7
------ = ------
3x + 4 6x - 2</p>

<p>If x satisfies the equation above, then what is the value of x ?
(A) −2
(B) −1
(C) 0
(D) 1
(E) 2</p>

<p>I started putting in the answers till I got 2 but I might not be so lucky next time.</p>

<p>Would checking answers be the best approach?</p>

<p>are you asking how to solve the problem 3x+4 = 6x-2, what is x??</p>

<p>7 divided by (3x + 4) is equal to 7 divided by (6x-2)
is the same as solving 3x+4 = 6x-2 </p>

<p>6 = 3x
x = 2</p>

<p>Oh thanks, I didn’t see it that way since the numerators are the same.</p>

<p>TY :)</p>

<p>When each side of a particular square is lengthened by 2 inches, the area of the square
increases by 32 square inches. What is the length in inches of a side of the original
square?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8</p>

<p>How would you solve this?</p>

<p>Simple algebra, let x = the original side.</p>

<p>x(squared) + 32 = (x+2)(x+2)
x(squared) + 32 = x(squared) + 4x + 4</p>

<p>Subtract x squared from both sides.</p>

<p>32 = 4x + 4</p>

<p>Subtract 4 from each side.</p>

<p>4x = 28
x = 7</p>

<p>To check: 7 x 7 = 49
9 x 9 = 81</p>

<p>81 - 49 = 32!</p>

<p>(D) 7</p>

<p>Thank you :slight_smile: You guys are amazing. Sorry if the questions I’m asking are easy, I’m trying to review math before I take my first practice test.</p>

<p>I have a few more questions but I’ll try to figure them out myself first. Algebra was just too long ago.</p>

<p>A triangle is to be changed by increasing the length of its base by 40% and decreasing
the length of its height by 40%. Which of the following must be true about the area
of the triangle?
(A) It will increase by 40%.
(B) It will increase by 16%.
(C) It will not change.
(D) It will decrease by 16%.
(E) It will decrease by 40%.</p>

<p>Any way to solve that quickly with algebra without having to test it with a hypothetical triangle?</p>

<p>You can solve it with the constants given in the triangular area formula.</p>

<p>Area of original triangle = .5<em>b</em>h = A</p>

<p>New triangle = .5<em>.6b</em>1.4h = .84A</p>

<p>Thus, the triangle’s area has been decreased by 16%. Hope this is clear; tell me if I should elaborate or rephrase.</p>

<p>I get that .6 * 1.4 is .84 but I don’t really understand the logic, I’ll keep thinking on it though :D</p>

<p>NVM I just realized it :smiley: Those are the only things that change so you’ll have .84 of the original area.</p>

<p>Alright, I’ll try to rephrase. Decreasing the base by 40% would give you a base that is only 60% of the original, therefore this can be written as .6b. Same deal with the height, except it’s increased. Now that you have the values of the new triangle, you plug it into the formula. </p>

<p>A = .5<em>.6b</em>1.4h</p>

<p>I factored out the .5<em>b</em>h, because those were the values of the original triangle, and substituted a singular variable A in its place. .6<em>1.4</em>A is equal to .84<em>A, and .84</em>A is 84% of A (if this doesn’t make sense, substitute a number such as 100 in the place of A). The original triangle is by definition 100% of A, so a drop from 100% to 84% is a decrease of 16% in the area of the triangle.</p>

<p>If you want to stay away from abstract approach (which is nonetheless fast and accurate,) you can quickly build a triangle with HELPFUL numbers. Since the problem is stated in percentage, build one that has an area of 100. </p>

<p>A super easy one would be to use a base of 20 and a height of 10. So far so good, the area is 100. Now change the numbers as suggested in the problem. Base becomes 28 and height 6. The area is thus 1/2 * 6 * 28 … or 84. Now compare 100 and 84, and that is a 16 reduction from 100. Or a 16 percent reduction. </p>

<p>Please note that the explanation is lenghty but practice makes those manipulations easy. After a while, it becomes second nature to plug in easy numbers for triangles.</p>

<p>New problem I need help with.</p>

<p>Square root(x^2 - t^2) = 2t - x</p>

<p>If x and t are positive numbers that satisfy the equation above, what is the value of x/t</p>

<p>I would believe that the long way is the only approach so this is how you do it</p>

<p>sqrt(x^2-t^2) = 2t-x
(sqrt(x^2-t^2))^2 = (2t-x)^2 square each side to remove the sqrt sign on the left side.
x^2-t^2 = (4t^2 - 4tx + x^2) expand the right side.
0 = 5t^2-4tx cancel the x^2 on both sides and add t^2 to both sides.
0 = t(5t-4x) factor.
t = 0 which cannot be true and then 5t-4x = 0
5t = 4x add 4x to both sides.
5 = 4(x/t) divide both sides by t.
5/4=x/t divide both sides by 4.</p>

<p>So answer is 5/4.</p>

<p>Ty, do you by any chance know any ways to kind of catch up on math? I have the blue book and it doens’t really show you very well how to solve things. Should I just start doing the practice tests and that’s how I will learn the math?</p>

<p>square both sides to get rid of the square root.
x^2-t^2 = 4t^2-4tx+x^2
4t^2-4tx = -t^2/4t
t-x=-1/4t
4t-4x=-t
5t=4x
x/t=5/4
;)</p>

<p>Brolex, the questions you are asking are from Algebra 1.</p>

<p>So my advice is start there and work your way up. Yes, you can buy the blue book but it looks like you need the basic algebra foundation first. Otherwise you’re going to be learning this piecemeal. That said, I haven’t used a prep book to learn algebra, so it might work.</p>

<p>I just saw that algebra was a long time ago for you. It was 21 years ago for me, last math class was 15 years ago, so I know how you feel. However… I still think you need a tutor. Because once you remember the concepts, it should all suddenly appear fairly do-able. If that is not happening you need more than superficial review.</p>

<p>I talked to my calc teacher today and he explained most of the stuff. I’m going to buy the Grubbers math book, I’ve heard that’s the best one at least. Any other books I should consider in addition?</p>

<p>Khan Academy has videos and exercises in Algebra, Geometry, and Algebra 2.
Here is the link:
[Khan</a> Academy](<a href=“Khan Academy | Free Online Courses, Lessons & Practice”>Khan Academy | Free Online Courses, Lessons & Practice)</p>

<p>Its free, so it might be a good resource. Another good strategy might be to obtain a textbook on Algebra and do a few problems from each chapter, especially harder ones. If you have access to one, IB Math Studies textbooks are pretty good for SAT Math review (if nothing else is available). They’re about equivalent to Algebra 2, Geometry, Trig, and some Stats. Its analogous to Pre-Calc, actually.
Getting a good (older ones usually have better problems, IMO) Algebra textbook plus using Khan Academy for explanation if necessary is probably a good strategy.</p>

<p>How can you be in calculus and not be able to do algebra? Isn’t algebra done every day in calculus classes? Or at least weekly?</p>

<p>I strongly recommend Saxon math for revising.</p>