{"id":12472,"date":"2020-09-23T02:16:30","date_gmt":"2020-09-22T18:16:30","guid":{"rendered":"https:\/\/www.intmath.com\/blog\/?p=12472"},"modified":"2020-09-23T02:16:30","modified_gmt":"2020-09-22T18:16:30","slug":"solving-systems-of-equations-by-using-elimination","status":"publish","type":"post","link":"https:\/\/www.intmath.com\/blog\/mathematics\/solving-systems-of-equations-by-using-elimination-12472","title":{"rendered":"Solving Systems of Equations by Using Elimination"},"content":{"rendered":"<p>In mathematics, an equation is a statement where two mathematical expressions are equal to each other. Because this is algebra, there must be a\u00a0<strong>variable<\/strong>\u00a0in the equation. A variable is an unknown number, and we end up mostly solving these variables to prove the equation true. In some cases, we'll have to solve an equation that uses more than one variable and one equation. There are plenty of established methods for solving these equations, but one of the more common ways is by using elimination.<\/p>\n<p>Let's first review some key points about equations.<\/p>\n<h2>Equation Review<\/h2>\n<p>An equal sign separates the two mathematical expressions of an algebraic equation. One expression is on the right-hand side of the equal sign, and the other expression is on the left-hand side of the equal sign.<\/p>\n<p>An example of an equation is:<\/p>\n<p><em>2x + 3 = 12<\/em><\/p>\n<p>The left-hand side, which is\u00a0<em>2x + 3,<\/em>\u00a0is equal to the right-hand side,\u00a0<em>12<\/em>.<\/p>\n<p>When dealing with equations, you'll often come across these other terms:<\/p>\n<ul>\n<li><strong>Term<\/strong>: the part(s) of the equation usually separated by a plus or minus sign. In the example given, the terms in the equation are\u00a0<em>2x, 3, and 12.<\/em><\/li>\n<li><strong>Coefficient:<\/strong>\u00a0a number associated with a variable by multiplication. In the example above, the coefficient of\u00a0<em>x<\/em>\u00a0is 2. Only variables have coefficients.<\/li>\n<li><strong>Constants<\/strong>: terms of an equation that do not have an associated variable.<\/li>\n<li>S<strong>olving the Equation<\/strong>: finding the values of the variables that make the equation true.<\/li>\n<\/ul>\n<h2>Elimination Method<\/h2>\n<p>Some equations are very simple, and you can\u00a0solve them\u00a0without needing elaborate methods, like\u00a0<em>y = 3\u00a0<\/em>or\u00a0<em>x + 1 = 3.<\/em><\/p>\n<p>However, some equations are complex and require an established method\u00a0for finding the solution. The\u00a0<strong>elimination method<\/strong>\u00a0is used for solving equations that have more than one variable and\u00a0more than one equation. In the elimination method, you eliminate one of the variables\u00a0to solve for the remaining one. Once you have solved for that variable's value, you can substitute the value\u00a0into any of the equations to find the other variable.<\/p>\n<p>Generally, if an equation contains two unknown variables, you need at least two equations to solve for the two unknown variables. This is called\u00a0<strong>system equations<\/strong>.<\/p>\n<p><em>2x + 3y = 7 \u2026. eqn 1<\/em><\/p>\n<p><em>x \u2013 y = 2 \u2026. eqn 2<\/em><\/p>\n<p>Eqn 1 and Eqn 2 form\u00a0a system equation. The two unknown variables in the two equations are\u00a0<em>x\u00a0<\/em>and<em>\u00a0y.<\/em><\/p>\n<h3>Example 1: Two Equations<\/h3>\n<p>Solve the system equation below using the elimination method.<\/p>\n<p><em>3x \u2013 2y = 9 \u2026.. (eqn 1)<\/em><\/p>\n<p><em>6x \u2013 y = 27 \u2026.. (eqn 2)<\/em><\/p>\n<p>Let\u2019s call the first equation Eqn 1 and the second equation Eqn 2.<\/p>\n<h4><strong>Step 1<\/strong><\/h4>\n<p>The first step is to choose which variable to eliminate. I am going to eliminate\u00a0<em>x<\/em>. Before you can eliminate, the coefficients of the variable in the two equations must be the same. The coefficient of\u00a0<em>x<\/em>\u00a0in eqn 1 must be the same as the coefficient of\u00a0<em>x<\/em>\u00a0in eqn 2. But they are not the same, so we have to make them the same. You can change the coefficients of variables by multiplying the equation with constants. So let's multiply eqn 1 by 2.<\/p>\n<p><em>2 * (3x \u2013 2y = 9)<\/em><\/p>\n<p><em>6x \u2013 4y = 18 \u2026. (eqn 3)<\/em><\/p>\n<p>As you can see, we multiplied all the terms of the equation by 2. You can also choose to divide an equation by a constant if you prefer.<\/p>\n<p>Our eqn 1 is now eqn 3.<\/p>\n<h4><strong>Step 2<\/strong><\/h4>\n<p>Since the coefficients of\u00a0<em>x<\/em>\u00a0are now the same, we can proceed with the elimination.<\/p>\n<p>Let's subtract eqn 3 from eqn 2.<\/p>\n<p><em>\u00a0 \u00a0 \u00a06x \u2013 y = 27<\/em><\/p>\n<p><em>\u00a0 \u2013 6x \u2013 4y = 18 \u00a0\u00a0\u00a0\u00a0<\/em><\/p>\n<p><em>\u00a0 \u00a0 0\u00a0 + 3y = 9\u00a0 \u2026\u2026 (eqn 4)<\/em><\/p>\n<p>A few notes on this subtraction:<\/p>\n<ul>\n<li>The coefficient of\u00a0<em>x\u00a0<\/em>is now<em>\u00a00<\/em>.<\/li>\n<li><em>-y \u2013 (- 4y) = -y + 4y = 3y<\/em><\/li>\n<\/ul>\n<p>We now have eqn 4, which is\u00a0<em>3y = 9.<\/em><\/p>\n<h4><strong>Step 3<\/strong><\/h4>\n<p>Solve the resulting equation to find the remaining variable.<\/p>\n<p><em>3y = 9<\/em><\/p>\n<p><em>y = 9 \/ 3<\/em><\/p>\n<p><em>y = 3.<\/em><\/p>\n<p>We now know the value of\u00a0<em>y<\/em>\u00a0is 3.<\/p>\n<h4><strong>Step 4<\/strong><\/h4>\n<p>Substitute the value of\u00a0<em>y = 3<\/em>\u00a0into eqn 2 to find the value of\u00a0<em>x<\/em>.<\/p>\n<p><em>6x \u2013 y = 27<\/em><\/p>\n<p>Replace\u00a0<em>y<\/em>\u00a0with\u00a0<em>3.<\/em><\/p>\n<p><em>6x \u2013 3 = 27<\/em><\/p>\n<p><em>6x = 30<\/em><\/p>\n<p><em>x = 30 \/ 6<\/em><\/p>\n<p><em>x = 5.<\/em><\/p>\n<p>We have solved the system of equations to arrive at\u00a0<em>x = 5 and y = 3<\/em>.<\/p>\n<p>Let us look at another example.<\/p>\n<h3>Example 2: Three Equations<\/h3>\n<p>Solve the systems of equations below.<\/p>\n<p><em>2x \u2013 y + 3z = 14 .... (eqn 1)<\/em><\/p>\n<p><em>x + 5y \u2013 2z = 20 \u2026. (eqn 2)<\/em><\/p>\n<p><em>x \u2013 y = 4 \u2026. (eqn 3)<\/em><\/p>\n<p>The above system equations contain three variables<em>\u00a0x, y,\u00a0<\/em>and<em>\u00a0z<\/em>. The third equation does not have the\u00a0<em>z<\/em>\u00a0variable. This only means that the coefficient of\u00a0<em>z<\/em>\u00a0in eqn 3 is 0. So if you are to subtract, you will simply include\u00a0<em>0z<\/em>\u00a0in eqn 3.<\/p>\n<h4><strong>Step 1<\/strong><\/h4>\n<p>By looking at the three equations, subtracting\u00a0any two equations won't leave us with only one variable, because there are three variables. If we eliminate one, we still have two variables left. There is something else we can do, though.<\/p>\n<p>Consider eqn 3. It has only two variables, but we can express\u00a0<em>y<\/em>\u00a0in terms of\u00a0<em>x<\/em>.<\/p>\n<p><em>x \u2013 y = 4<\/em><\/p>\n<p><em>x = 4 + y \u2026. (eqn 4)<\/em><\/p>\n<p>By moving\u00a0<em>y\u00a0<\/em>to the right side of the equation, we have a new equation to help us solve the problem.<\/p>\n<h4><strong>Step 2<\/strong><\/h4>\n<p>Substitute eqn 4 into eqn 1. This means we will\u00a0replace the\u00a0<em>x<\/em>\u00a0in eqn 1 with\u00a0<em>4 + y<\/em><\/p>\n<p><em>2x \u2013 y + 3z = 14<\/em><\/p>\n<p><em>2 (4 + y) \u2013 y + 3z = 14<\/em><\/p>\n<p><em>8 + 2y \u2013 y + 3z = 14<\/em><\/p>\n<p><em>y + 3z = 14 \u2013 8<\/em><\/p>\n<p><em>y + 3z = 6 \u2026. (eqn 5)<\/em><\/p>\n<p>We also have a new equation.<\/p>\n<h4><strong>Step 3<\/strong><\/h4>\n<p>Substitute eqn 4 into eqn 2.<\/p>\n<p><em>x + 5y \u2013 2z = 20<\/em><\/p>\n<p><em>4 + y + 5y \u2013 2z = 20<\/em><\/p>\n<p><em>6y \u2013 2z = 20 \u2013 4<\/em><\/p>\n<p><em>6y \u2013 2z = 16 \u2026.. (eqn 6)<\/em><\/p>\n<p>This makes eqn 6, where there are now two\u00a0variables.<\/p>\n<h4><strong>Step 4<\/strong><\/h4>\n<p>The next step is to eliminate\u00a0<em>y<\/em>. But we first need to make the coefficient of\u00a0<em>y<\/em>\u00a0in eqn 5 the same as in eqn 6. So we multiply eqn 5 by 6.<\/p>\n<p><em>6 * (y + 3z = 6)<\/em><\/p>\n<p><em>6y + 18z = 36 \u2026. (eqn 7)<\/em><\/p>\n<p>We can now subtract eqn 6 from eqn 7.<\/p>\n<p><em>\u00a0 \u00a0 6y + 18z = 36<\/em><\/p>\n<p><em>\u2013 (6y \u2013 2z = 16)\u00a0<\/em><\/p>\n<p><em>0 + 20z = 20 \u2026. (eqn 8)<\/em><\/p>\n<p>Eqn 8 only contains one variable.<\/p>\n<h4><strong>Step 5<\/strong><\/h4>\n<p>Solve for\u00a0<em>z<\/em>.<\/p>\n<p><em>20z = 20<\/em><\/p>\n<p><em>z = 1<\/em><\/p>\n<p>Now substitute\u00a0<em>z = 1<\/em>\u00a0into eqn 5.<\/p>\n<p><em>y + 3z = 6<\/em><\/p>\n<p><em>y + 3 * 1 = 6<\/em><\/p>\n<p><em>y = 3.<\/em><\/p>\n<p>Now substitute\u00a0<em>y = 3<\/em>\u00a0into eqn 4.<\/p>\n<p><em>x = 4 + y<\/em><\/p>\n<p><em>x = 4 + 3<\/em><\/p>\n<p><em>x = 7.<\/em><\/p>\n<p>The solution to the system equations is\u00a0<em>x = 7, y = 3 and z = 1<\/em>.<\/p>\n<p>The elimination method is not difficult to learn, but you must stay organized. Variables and substitutions can get pretty messy and confusing if you don't lay them out on the paper correctly. And, as you can see, some equations take more than a few steps to complete. Just keep your pencil handy and have plenty of scrap paper to show your work.<\/p>\n<p class=\"alt\"><a href=\"#respond\" id=\"comms\">Be the first to comment<\/a> below.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In mathematics, an equation is a statement where two mathematical expressions are equal to each other. Because this is algebra, there must be a\u00a0variable\u00a0in the equation. A variable is an unknown number, and we end up mostly solving these variables to prove the equation true. In some cases, we'll have to solve an equation that [&hellip;]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mo_disable_npp":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12472"}],"collection":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/comments?post=12472"}],"version-history":[{"count":1,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12472\/revisions"}],"predecessor-version":[{"id":12473,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12472\/revisions\/12473"}],"wp:attachment":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/media?parent=12472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/categories?post=12472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/tags?post=12472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}