{"id":13491,"date":"2017-02-15T06:38:07","date_gmt":"2017-02-15T11:38:07","guid":{"rendered":"http:\/\/grockit.com\/gmat\/?p=1211"},"modified":"2020-09-11T20:42:25","modified_gmt":"2020-09-11T20:42:25","slug":"gmat-quantitative-solving-linear-equations-with-unknowns","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/gmat\/gmat-quantitative-solving-linear-equations-with-unknowns\/","title":{"rendered":"GMAT Quantitative: Solving Linear Equations with Unknowns"},"content":{"rendered":"

A linear equation is any equation where the highest power of the unknown, which I shall call x, is 1.\u00a0 To illustrate more clearly with a few examples:
\nx+1 = 4
\n10x = 3
\nx = 18 \u2013 4x
\nx2<\/sup> + 2 = 2x and x3<\/sup> = 8 are not linear equations because there are x\u2019s that are raised to a higher power than 1.
\n 
\n

Solving Equations with 1 Unknown<\/h3>
<\/div><\/div><\/div>
\nA linear equation with 1 variable is the simplest type to solve. There is 1 equation and 1 unknown, which means that the unknown can always<\/strong> be determined. To solve such an equation, you need to rearrange the equation to have like terms on either side of the equal sign. Put another way, you are trying to isolate x (or whatever the variable is called) on one side of the equation.
\nFor example, if 2x = 234, to isolate x, we have to divide the entire equation by 2. Doing this, we get x = 117.
\nIf there are x\u2019s and numbers on either side of the equal sign, we add and subtract values to isolate x on one side. Suppose 2x \u2013 17 = 18 \u2013 3x
\nThe first thing we could do is to add 17 to both sides to get: 2x \u2013 17 + 17 = 18 \u2013 3x + 17
\nThis reduces to 2x = 35 \u2013 3x
\nNow, we need to have all the x\u2019s on one side so we add 3x to both sides to get: 2x + 3x = 35 \u2013 3x + 3x
\nThis reduces to 5x = 35
\nDividing by 5 on both sides, we get x = 7
\n

Practice Question 2<\/h3>
<\/div><\/div><\/div>
\nWhat\u2019s a little trickier than manipulating algebraic equations is translating a word problem into an algebraic equation. Here\u2019s another practice problem:
\nJack and his brother are sharing a monster piece of licorice that is 28 inches long. Since Jack is older, his share is 8 inches longer than his brother’s. How long, in inches, is Jack’s brother’s piece?
\nThe way to solve this problem is to let something be x. Here\u2019s what happens if we let Jack\u2019s piece be x inches.
\nJack\u2019s piece = x inches
\nJack\u2019s brother\u2019s piece = x \u2013 8 inches
\nTotal length of licorice = Jack\u2019s piece + Jack\u2019s brother\u2019s piece = 28 = x + (x-8)
\nThis means that x = 18 inches. But remember that the question wants the length of Jack\u2019s brother\u2019s piece, which we have defined as x \u2013 8. So the correct answer is 10 inches.
\nHere\u2019s what happens if we let Jack\u2019s brother\u2019s piece be x inches.
\nJack\u2019s brother\u2019s piece = x inches
\nJack\u2019s piece = x + 8 inches
\nTotal length of licorice = 28 = x + (x+8) and we determine that x = 10. In this case, since we have already defined Jack\u2019s brother\u2019s piece to be x, there is no further step we need to take.
\nIn general, here are a few things to keep in mind:<\/p>\n
    \n
  • if there is only one unknown, you only need one equation to determine the value of the unknown<\/li>\n
  • in dealing with algebraic equations, remember that anything you do to one side (be it adding, subtracting, multiplying or dividing) you need to do to this other side too.<\/li>\n
  • in dealing with word problems, define something to be x and see if you can define other things in terms of x only. (For example, in the question about licorice, you would not want to let Jack\u2019s piece be x inches and his brother\u2019s be y inches.) Don\u2019t introduce unnecessary variables if it can be expressed in terms of an existing variable.<\/li>\n<\/ul>\n

    Solving Equations with 2 Unknown<\/h3>
    <\/div><\/div><\/div>
    \nNow that we’ve covered linear equations with one unknown, we can move on to tackle equations with two unknowns. In order to solve such equations, you need at least 2 distinct<\/strong> equations involving these unknowns.
    \nFor example, if we are trying to solve for x and y, we won’t be able to solve it using these 2 equations.
    \n2x + y = 14
    \n4x + y – 14 = 14 – y
    \nWhy? Because the two equations on top are the same. If you simplify the second equation, you get 4x + 2y = 28 which reduces to 2x + y = 14 – the same equation as the first. If the two equations are the same, then there will be infinitely many values for x and y that will satisfy the equations. For example, x = 2 and y = 10 satisfies the equation. So does x = 4 and y = 8.\u00a0 And so does x = 6 and y = 2.
    \nIn order to solve for an actual value of x and y, we need 2 distinct<\/strong> equations.
    \nFor example, if we had
    \n2x + y = 14 ——–(1)
    \nx – y = 4 ———–(2)
    \nThen from equation (2), we can get x = 4 + y and substitute that into equation (1) to get:
    \n2(4 + y) + y = 14\u00a0 We can then solve for y. See if you got y = 2\u00a0 Once you’ve got y = 2, you can substitute that into x= 4 + y to get x = 6.
    \nAn important lesson here is that you need as many distinct equations as you have variables. So if you are doing a data sufficiency question, you can sometimes just note that as long as you have two distinct equations, you will be able to solve for x and y. Disregard or other equivalent equations.
    \nSometimes, however, its possible for there to be no solution<\/strong> to the set of equations. This occurs when one side of each equation is the same, but the other side is different. For example,
    \n2x + y = 14
    \n2x + y = 0
    \nhas no solution. Its like saying that the same steak costs $14 and $0 at the same restaurant. Of course, the GMAT is not going to make it so obvious that the equations contradict each other. Usually, you would have to simplify one of the equations to check if it is
    \n1. the same as the other equation, in which case, you have infinitely <\/strong>many solutions
    \n2. the same as the other equation on one side, but different on the other side of the equal sign. In this case, you have no solution
    \n3. a distinct equation, in which case x and y has an exact value.
    \n

    Multiple Unknowns Practice Question<\/h3>
    <\/div><\/div><\/div>
    \nThe GMAT might also require you to translate a word problem into a pair of simultaneous equations to solve. Let’s try a practice question.
    \nA package contains nothing but 35 DVDs and 15 videotapes. What is the total weight, in pounds, of the contents of the package?
    \n(1) Each videotape weighs twice as much as each DVD.
    \nIf we let the weight of a DVD be x and a videotape be y, then according to this statement y = 2x
    \n(2) The total weight of 2 of the videotapes and 2 of the DVDs is 1 pound.
    \nFollowing the previous notation, 2y + 2x = 1
    \nWe clearly have 2 distinct equations meaning that we need both statements together to solve this question. The more you practice, the more\u00a0you’ll start to get the hang of constructing and simplifying equations to see if they are distinct and, thus, solvable.<\/p>\n","protected":false},"excerpt":{"rendered":"

    A linear equation is any equation where the highest power of the unknown, which I shall call x, is 1.\u00a0 To illustrate more clearly with a few examples: x+1 = 4 10x = 3 x = 18 \u2013 4x x2 + 2 = 2x and x3 = 8 are not linear equations because there are […]<\/p>\n","protected":false},"author":1,"featured_media":28713,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[55],"tags":[56,80],"_links":{"self":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/13491"}],"collection":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/comments?post=13491"}],"version-history":[{"count":1,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/13491\/revisions"}],"predecessor-version":[{"id":36011,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/13491\/revisions\/36011"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media\/28713"}],"wp:attachment":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media?parent=13491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/categories?post=13491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/tags?post=13491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}