{"id":1787,"date":"2019-08-24T12:01:25","date_gmt":"2019-08-24T17:01:25","guid":{"rendered":"http:\/\/www.kaptest.com\/blog\/prep\/?p=1787"},"modified":"2020-09-11T20:41:18","modified_gmt":"2020-09-11T20:41:18","slug":"psat-systems-of-equations-combinationsubstitution","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/psat\/psat-systems-of-equations-combinationsubstitution\/","title":{"rendered":"PSAT Systems of Equations: Combination\/Substitution"},"content":{"rendered":"

Now that you understand the requirements that must be satisfied\u00a0to solve a system of equations, let’s look at some methods for solving these systems effectively. The two main methods for solving a system of linear equations are substitution and combination (sometimes referred to as\u00a0elimination by addition<\/i>).
\nSubstitution<\/b> is the most straightforward method for solving systems, and it can be applied in every situation. Unfortunately, it is often the longest and most time-consuming route for solving systems of equations as well. To use substitution, solve the simpler\u00a0of the two equations for one variable, and then substitute the result into the other equation. You could use substitution to answer the following question, but you\u2019ll see that there\u2019s a quicker way: combination.
\nCombination<\/b> involves adding the two equations together to eliminate a variable. Often, one or both of the equations must be multiplied by a constant before they are added together. Combination is often the best technique to use to solve a system of equations as it is usually faster than substitution.
\nUnfortunately, even though most students prefer substitution, problems on the PSAT are often designed to be quickly solved with combination. To really boost your score on Test Day, practice combination as much as you can on Practice Tests\u00a0and in homework problems so that it becomes second nature.
\n\t

\t\t

\n1. If 6a<\/i> + 6b<\/i> = 30 and 3a<\/i> + 2b<\/i> = 14, then what are the values of a<\/i> and b\u00a0<\/i>?<\/p>\n

    \n
  1. <\/i>a<\/i> = 2; b<\/i> = 2<\/li>\n
  2. <\/i>a<\/i> = 4; b<\/i> = 1<\/li>\n
  3. <\/i>a<\/i> = 1; b<\/i> = 4<\/li>\n
  4. <\/i>a<\/i> = 3; b<\/i> = 1<\/li>\n<\/ol>\n

    \n<\/div><\/div>
    \nWork through\u00a0the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan\u2019s strategic thinking on the left, along with suggested math scratchwork on the right.<\/p>\n\n\n\n\n\n\n
    Strategic Thinking<\/b><\/td>\nMath Scratchwork<\/b><\/td>\n<\/tr>\n
    Step 1: Read the question, identifying and organizing important information as you go<\/b>You are given a system of two equations with two unknowns and asked to find the values of a<\/i> and b<\/i>.<\/td>\n6a<\/i> + 6b<\/i> = 303a<\/i> + 2b<\/i> = 14<\/td>\n<\/tr>\n
    Step 2: Choose the best strategy to answer the question<\/b>Remember, while substitution could be used to solve this type of problem, combination will often be faster. What transformation will enable you to add the equations and eliminate a variable?<\/i>Combination often requires you to multiply one of your equations by a constant. In this case, notice what happens if you multiply the second equation by \u20133.
    \nWhat\u2019s the next step in combination?<\/i>
    \nBy arranging the equations vertically, you can simply add them, combining like terms along the way.
    \nNotice that 6b<\/i> + (\u20136b<\/i>) = 0b<\/i> = 0, and you\u2019ve eliminated b<\/i> from your equation. Your goal when using combination is to set the coefficient of the variable you are trying to eliminate to a number that is equal in magnitude and opposite in sign to the coefficient in the other equation. Now you can easily solve for a<\/i>.<\/td>\n    		[raw](\u22123)(3a +2b) = (14)(\u22123)
    \n\u22129a \u2212 6b = \u221242
    \n6a\u00a0+\u00a06b\u00a0=\u00a030+\u00a0\u00a0\u00a0\u2212\u00a09a\u00a0\u2212\u00a06b\u00a0=\u00a0\u221242\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u22123a\u00a0+\u00a00b\u00a0=\u00a0\u221212 [\/raw]\u22123a<\/em>\u00a0 + \u00a00b =\u00a0<\/em>\u221212
    \n\u22123a<\/i> = \u221212
    \na<\/i>\u00a0=\u00a04<\/td>\n<\/tr>\n
    Step 3: Check that you answered the<\/b> right<\/i><\/b> question<\/b>Even though the question asks you for the values of a<\/i> and b<\/i>, each answer choice has a different value of a<\/i>. There\u2019s no need to plug back in and find the correct value of b<\/i>. Choice (B) is correct.<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    PSAT System of Equations: Combination Strategy<\/h3>
    <\/div><\/div><\/div>
    \nCombination can also be used when the test makers ask you for a strange quantity, as in the following problem:
    \n\t
    \t\t
    <\/p>\n
    \n

    If\u00a05c<\/i>\u00a0\u2212 2b<\/i>\u00a0= 15<\/span>\u00a0and\u00a03b<\/i>\u00a0\u2212 4c<\/i>\u00a0= 12,<\/span>\u00a0what is the value of\u00a0b<\/i>\u00a0+\u00a0c\u00a0<\/i>?<\/span><\/p>\n<\/section>\n

      \n
    1. \u221227<\/li>\n
    2. \u00a0 \u22123<\/li>\n
    3. \u00a0 \u00a0 8<\/li>\n
    4. \u00a0 27<\/li>\n<\/ol>\n

      \n<\/div><\/div>
      \n

      <\/span><\/span><\/div><\/p>\n\n\n\n\n\n\n
      Strategic Thinking<\/b><\/td>\nMath Scratchwork<\/b><\/td>\n<\/tr>\n
      Step 1: Read the question, identifying and organizing important information as you go<\/b>You are being asked to find the value of b<\/i> + c<\/i>. The question stem provides two equations involving b<\/i> and c<\/i>.<\/td>\n5c<\/i> \u2013 2b<\/i> = 15
      \n3b<\/i> \u2013 4c<\/i> = 12<\/td>\n<\/tr>\n
      Step 2: Choose the best strategy to answer the question<\/b>How can you quickly and accurately answer the question? Why are the test makers asking for the quantity
      \nb +<\/i> c and not the values of<\/i> b and<\/i> c independently?<\/i>The fact that you\u2019re solving for b<\/i> + c<\/i> suggests that there\u2019s a time-saving shortcut to be found. Because you\u2019re not trying to get rid of a variable, see if you can add the equations to get a result that has b<\/i> + c<\/i> equal to some numerical value.\u00a0Before you add, don’t forget to write the variable terms in the same order for each equation.<\/td>\n      		[raw]\u00a0\u00a0\u00a0\u00a0\u00a0\u22122b\u00a0+\u00a05c\u00a0=\u00a015+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03b\u00a0\u2212\u00a04c\u00a0=\u00a012\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0b\u00a0\u00a0+\u00a0\u00a0c\u00a0=\u00a027
      \n[\/raw]<\/td>\n<\/tr>\n
      Step 3: Check that you answered the<\/b> right<\/i><\/b> question<\/b>Because you\u2019re asked to find value of b<\/i> + c,<\/i> there\u2019s nothing more to do here. Choice (D) is correct.<\/td>\nb<\/i> + c<\/i> = 27<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      That was\u00a0much easier and faster than substitution. With substitution, you could spend more than\u00a0two minutes solving a question like this. However, a bit of analysis and combination gets the job done in much less time.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Now that you understand the requirements that must be satisfied\u00a0to solve a system of equations, let’s look at some methods for solving these systems effectively. The two main methods for solving a system of linear equations are substitution and combination (sometimes referred to as\u00a0elimination by addition). Substitution is the most straightforward method for solving systems, […]<\/p>\n","protected":false},"author":1,"featured_media":28745,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[240],"tags":[],"_links":{"self":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/1787"}],"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=1787"}],"version-history":[{"count":3,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/1787\/revisions"}],"predecessor-version":[{"id":34803,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/1787\/revisions\/34803"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media\/28745"}],"wp:attachment":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media?parent=1787"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/categories?post=1787"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/tags?post=1787"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}