{"id":12267,"date":"2016-11-22T06:00:06","date_gmt":"2016-11-22T11:00:06","guid":{"rendered":"http:\/\/grockit.com\/blog\/collegeprep\/?p=2194"},"modified":"2020-09-11T20:42:32","modified_gmt":"2020-09-11T20:42:32","slug":"act-math-systems-of-linear-equations","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/act\/act-math-systems-of-linear-equations\/","title":{"rendered":"ACT Math: Systems of Linear Equations"},"content":{"rendered":"<p>A system of equations is a set of two or more equations that have two or more variables. \u201cSolving\u201d the system of equations means finding values for each of the variables that make each equation true. You probably remember from school that there are a few different ways to solve a system of equations. This shouldn\u2019t deter you. While it takes some practice to choose the quickest method for solving a particular system of equations, you\u2019ll eventually find that having options greatly increases your speed on the test. Let\u2019s check out some methods.<br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-0  el_before_av_heading  avia-builder-el-first  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Addition or Subtraction<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nIn some rare cases, you can simply add or subtract both equations to isolate a variable. If the equations are set up the right way, addition\/subtraction is by far the quickest choice.<br \/>\nSolve:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x+y=10<br \/>\nx \u2013y= 2<br \/>\nLet\u2019s add:\u00a0\u00a0 2x= 12\u2192 x=6<br \/>\nPlug in 6 for x: 6+y=10\u2192 y=4<br \/>\nNotice how fast that was. Because the y\u2019s canceled upon addition, we quickly isolated the x variables in order to solve for x. If one of the y\u2019s had a coefficient of, say, 7, then we could have chosen to subtract the equations and thus cancel the x\u2019s.<br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-1  el_after_av_heading  el_before_av_heading  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Substitution<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nSubstitution might be the method you\u2019re most comfortable with. It\u2019s called \u2018substitution\u2019 because you are substituting a variable in one equation (say, x) for an expression derived from the other equation (say, y+5); in this way, you write one equation in terms of only one variable, rendering it solvable.<br \/>\nSolve: 2x+ y = 10<br \/>\n-6x + 2y = 12<br \/>\nLet\u2019s solve for y in the first equation:2x+y=10\u2192 y=10-2x<br \/>\nNow, let\u2019s substitute 10-2x for y in the other equation: -6x+2(10-2x)=12\u2192 -6x+20-4x=12<br \/>\n\u2192 -10x+20=12\u00e810x=8 \u2192 x=8\/10\u2192 <strong>x=4\/5<\/strong><br \/>\nNow, let\u2019s plug in 4\/5 for x in the first equation (because it\u2019s simpler): 2(4\/5)+y=10\u2192 8\/5+y=10<br \/>\n\u2192 y=10- (8\/5)\u2192 y=50\/5\u00a0 &#8211;\u00a0 8\/5 \u2192 y=<strong>42\/5<\/strong><br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-2  el_after_av_heading  el_before_av_heading  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Elimination<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nElimination is essentially solving by addition\/subtraction, but it requires a little more manipulation. Basically, elimination is manipulating a system of equations so that the addition or subtraction will become a viable method.<br \/>\nSolve:\u00a0\u00a0\u00a0 x+2y = 10<br \/>\n4x+y\u00a0 =\u00a0 8<br \/>\nBasically, we can multiply either of the equations by some constant in order to make the variables cancel upon addition or subtraction. In this case, we can either multiply the top equation by -4 (to cancel the x\u2019s) or multiply the second equation by -2 (to cancel the y\u2019s); let\u2019s go for the second option.<br \/>\nX+2y=10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x+2y=10<br \/>\n-2(4x+y=8)\u2192\u00a0\u00a0\u00a0\u00a0 -8x-2y=-16<br \/>\nNow, add: -7x=-6\u2192 x=6\/7<br \/>\nLastly, substitute 6\/7 for x in one of the original equations. I\u2019ll choose the first:<br \/>\n6\/7\u00a0\u00a0 + 2y=10 \u2192 2y=70\/7\u00a0 &#8211;\u00a0 6\/7 \u2192 2y=\u00a0 64\/7 \u2192\u00a0 y=64\/7 *\u00a0 \u00bd\u00a0 \u2192 y=64\/14\u00a0\u2192 <strong>y=32\/7<\/strong><br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-3  el_after_av_heading  avia-builder-el-last  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Tricky Problems<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nSome systems of equations problems will not ask you to solve, but to perform some other calculation. Let\u2019s see an example:<br \/>\nIf 3a+5b=10 and 5a+3b=30, what is the average of a and b?<br \/>\nFirst, set them up:\u00a0 3a+5b=10<br \/>\n5a+ 3b=30<br \/>\nNext, consider what average means. The average of two variables is their sum (a+b) divided by 2 (the number of items). So, what we really want to solve for is a+b. We could get at this by using elimination and substitution, but look how fast we can solve using addition.<br \/>\n3a+5b = 10<br \/>\n+ <span style=\"text-decoration: underline\">5a + 3b = 30<\/span><br \/>\n8a + 8b = 40<br \/>\nNext, divide by 8 to yield a+b:<br \/>\na+b = 5<br \/>\nIf a+b is 5, then the average of a and b is 5\/2.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A system of equations is a set of two or more equations that have two or more variables. \u201cSolving\u201d the system of equations means finding values for each of the variables that make each equation true. You probably remember from school that there are a few different ways to solve a system of equations. This [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":27159,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[58],"tags":[792],"_links":{"self":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/12267"}],"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=12267"}],"version-history":[{"count":3,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/12267\/revisions"}],"predecessor-version":[{"id":36140,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/12267\/revisions\/36140"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media\/27159"}],"wp:attachment":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media?parent=12267"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/categories?post=12267"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/tags?post=12267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}