{"id":1592,"date":"2012-09-13T23:45:09","date_gmt":"2012-09-13T23:45:09","guid":{"rendered":"http:\/\/testpreparations.com\/help\/?p=1592"},"modified":"2014-12-02T08:32:02","modified_gmt":"2014-12-02T08:32:02","slug":"systems-of-linear-equations","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/help\/systems-of-linear-equations\/","title":{"rendered":"Systems of Linear Equations"},"content":{"rendered":"<p>A system of linear equations is a set of linear equations that have common variables. Common systems consist of two variables, x and y, and two linear equations. The solution to the system is the value of x and y that satisfy both equations. There are two ways to solve systems:<strong> substitution <\/strong>and<strong> elimination.<\/strong><\/p>\n<p>We will solve the following system in both ways.<\/p>\n<p>1) x + y = 3<\/p>\n<p>2) -3x + 5y = -1<\/p>\n<h5>Substitution<\/h5>\n<p><strong>Step 1: <\/strong>Solve one equation in terms of one of the variables whichever is easier.<\/p>\n<p>We rearrange equation 1 to form y = -x + 3.<\/p>\n<p><strong>Step 2: <\/strong>Substitute one equation <strong>into<\/strong> the other in terms of the variable found in step one.<\/p>\n<p>-3x + 5(-x + 3) = -1<\/p>\n<p><strong>Step 3:<\/strong> Solve for the variable.<\/p>\n<p>-3x &#8211; 5x + 15 = -1<\/p>\n<p>-8x = -16<\/p>\n<p>x = 2<\/p>\n<p>Step 4: Substitute the value found back into one of the <span style=\"text-decoration: underline\">original <\/span>equations and solve for the other variable.<\/p>\n<p>x + y = 3<\/p>\n<p>2 + y = 3<\/p>\n<p>y = 1<\/p>\n<p>So the solution to the system is x = 2 and y = 1.<\/p>\n<h5>Elimination<\/h5>\n<p><strong>Step 1: <\/strong>Multiple one or both equations by a value so that the absolute value of one variable&#8217;s coefficients are the same.<\/p>\n<p>Multiply (x + y = 3) by 3 to get 3x + 3x = 9<\/p>\n<p>Now our equations are<\/p>\n<p>1) 3x + 3y = 9<\/p>\n<p>2) -3x + 5y = -1<\/p>\n<p><strong>Step 2: <\/strong>Add or subtract the sides of the equation to each other, whichever would eliminate the one variable with the same coefficients.<\/p>\n<p>We are adding so that we can eliminate x.<\/p>\n<p>3x + 3y + (-3x + 5y) = 9 + (-1)<\/p>\n<p><strong>Step 3:<\/strong> Simplify and solve for the remaining variable.<\/p>\n<p>8y = 8<\/p>\n<p>y = 1<\/p>\n<p><strong>Step 4: <\/strong>Substitute the found value back into the one of the <span style=\"text-decoration: underline\">original<\/span> equations and solve for the other variable.<\/p>\n<p>x + y = 3<\/p>\n<p>x + 1 = 3<\/p>\n<p>x = 2<\/p>\n<p>So the solution to the system is x = 2 and y = 1.<\/p>\n<p>&nbsp;<\/p>\n<p>This article was written for you by <strong>Jeremie<\/strong>, one of the tutors with <span class=\"tutorOrange\">Test Prep Academy.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>A system of linear equations is a set of linear equations that have common variables. Common systems consist of two variables, x and y, and two linear equations. The solution to the system is the value of x and y that satisfy both equations. There are two ways to solve systems: substitution and elimination. We [&hellip;]<\/p>\n","protected":false},"author":6,"featured_media":2186,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[2841,3015,3021,2851],"tags":[75,2822,1752,3310,3322],"class_list":["post-1592","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-act","category-math-act","category-mathematics-sat","category-sat","tag-algebra-2","tag-elimination","tag-substitution","tag-system-of-equations","tag-two-variables"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/1592","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/comments?post=1592"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/1592\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media\/2186"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media?parent=1592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/categories?post=1592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/tags?post=1592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}