{"id":12425,"date":"2020-07-03T21:31:38","date_gmt":"2020-07-03T13:31:38","guid":{"rendered":"https:\/\/www.intmath.com\/blog\/?p=12425"},"modified":"2020-07-03T21:31:38","modified_gmt":"2020-07-03T13:31:38","slug":"solving-equations-with-the-addition-method","status":"publish","type":"post","link":"https:\/\/www.intmath.com\/blog\/mathematics\/solving-equations-with-the-addition-method-12425","title":{"rendered":"Solving Equations With The Addition Method"},"content":{"rendered":"

There are two ways of solving an equation: the addition method and the substitution method. Here, we\u2019re going to take a look at the addition method. The\u00a0addition method<\/strong>\u00a0of solving equations is an easy and simple\u00a0method used\u00a0in algebra. It\u2019s also called the\u00a0elimination method<\/strong>.<\/p>\n

Understanding the Parts of an Equation<\/h2>\n

Before we learn how to use the addition method, let's breakdown of the parts of an equation.<\/p>\n

This is our example:<\/p>\n

3x + 4y = 10<\/em><\/p>\n

The numbers 3 and 4 are coefficients that multiply with the variables.<\/p>\n

x<\/em>\u00a0and\u00a0y<\/em>\u00a0are variables that can have different values depending upon the outcome.<\/p>\n

Solving Using the Addition Method<\/h2>\n

Now how would you solve this equation? Ideally, you could\u00a0pair up this equation with another one, let\u2019s say\u00a04x + 2y = 20<\/em>\u00a0and solve it. But because both equations have two different variables, there is no way to identify the values of the variables. You would have to tediously guess the value of\u00a0x<\/em>\u00a0and\u00a0y<\/em>, plot it in on a graph, and find the point where the two lines intersect, which will then give you the coordinates of\u00a0x<\/em>\u00a0and\u00a0y<\/em>. Even then, if you draw the line or plot the points incorrectly, you won\u2019t get the correct values of\u00a0x<\/em>\u00a0and\u00a0y<\/em>.<\/p>\n

To save us the hassle of going through that process, we use the\u00a0addition method<\/strong>\u00a0for solving equations that have one or two variables.<\/p>\n

The addition method has a series of steps that you must\u00a0follow to solve the equation\u00a0correctly.<\/p>\n

Step 1<\/strong><\/h3>\n

Put two given equations on top of each other and label them.<\/em><\/p>\n

You need to determine what\u00a0the two equations are, first. After that, we need to put them on top of each other like a standard addition problem. We will use these two equations as an example:<\/p>\n

2x +\u00a0<\/em>4y =\u00a0<\/em>10\u00a0\u00a0 equation (1)<\/p>\n

4x -\u00a0<\/em>2y =\u00a0<\/em>15 \u00a0\u00a0\u00a0equation (2)<\/p>\n

Step 2\u00a0<\/strong><\/h3>\n

Select\u00a0the variable that needs to be removed.<\/em><\/p>\n

We now have to select the variable to remove from the equation.\u00a0 Let's choose the\u00a0y<\/em>\u00a0variable.<\/p>\n

Step 3<\/strong><\/h3>\n

Select\u00a0the equation to convert the coefficient of the variable.<\/em><\/p>\n

We now need to select the equation in which we will change the coefficient of the selected variable. Depending on the given equation, you can choose one or both. We will proceed with equation (2).<\/p>\n

Step 4<\/strong><\/h3>\n

Change the coefficient of the equation.<\/em><\/p>\n

After selecting the equation, remove the variable\u00a0to change the coefficient into its opposite. We chose equation (2). Since the\u00a0y<\/em>\u00a0coefficient is already negative, we don\u2019t need to convert the sign. We will just multiply both sides of the equation by 2. We will then end up with a new equation (3).<\/p>\n

(4x<\/em>\u00a0\u2013 2y<\/em>) x 2 = 15 x 2<\/p>\n

8x<\/em>\u00a0\u2013 4y<\/em>\u00a0= 30\u00a0\u00a0 Equation (3)<\/p>\n

Step 5<\/strong><\/h3>\n

Add the two equations and substitute the result into the original equation.<\/em><\/p>\n

Once the new equations are formed, we need to add the two equations and remove the selected variable. It is illustrated below:<\/p>\n

2x<\/em>\u00a0+ 4y<\/em>\u00a0= 10\u00a0\u00a0\u00a0 Equation (1)<\/p>\n

8x<\/em>\u00a0\u2013 4y<\/em>\u00a0= 30\u00a0\u00a0\u00a0 Equation (3)<\/p>\n

Add equation (1) and (3)<\/p>\n

10x<\/em>\u00a0= 40<\/p>\n

Divide by 10 on both sides to isolate and solve for\u00a0x<\/em>.<\/p>\n

x<\/em>\u00a0= 4<\/p>\n

Now that we found the value of\u00a0x<\/em>, we need to substitute it into the original equation (1)<\/p>\n

2(4) + 4y<\/em>\u00a0= 10<\/p>\n

8 + 4y<\/em>\u00a0= 10<\/p>\n

Subtract 8 from both sides<\/p>\n

4y<\/em>\u00a0= 2<\/p>\n

Divide both sides by 4<\/p>\n

y<\/em>\u00a0= 0.5<\/p>\n

Now we have the value of the two variables\u00a0x<\/em>\u00a0= 4 and\u00a0y<\/em>\u00a0= 0.5. We are done solving this equation.<\/p>\n

Conclusion<\/strong><\/h2>\n

There you have it! A simple and effective way to learn the addition or elimination method to solve equations. Don\u2019t worry, it doesn\u2019t matter what two equations you get. Follow the steps above, and you will have no problem solving any two-variable equation question presented to you.<\/p>\n

See the 1 Comment<\/a> below.<\/p>\n","protected":false},"excerpt":{"rendered":"

There are two ways of solving an equation: the addition method and the substitution method. Here, we\u2019re going to take a look at the addition method. The\u00a0addition method\u00a0of solving equations is an easy and simple\u00a0method used\u00a0in algebra. It\u2019s also called the\u00a0elimination method. Understanding the Parts of an Equation Before we learn how to use the […]<\/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\/12425"}],"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=12425"}],"version-history":[{"count":1,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12425\/revisions"}],"predecessor-version":[{"id":12426,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12425\/revisions\/12426"}],"wp:attachment":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/media?parent=12425"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/categories?post=12425"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/tags?post=12425"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}