{"id":35227,"date":"2023-10-16T16:32:14","date_gmt":"2023-10-16T11:02:14","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=35227"},"modified":"2023-11-10T10:29:09","modified_gmt":"2023-11-10T04:59:09","slug":"solvability-of-linear-simultaneous-equations","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/solvability-of-linear-simultaneous-equations\/","title":{"rendered":"Solvability of Linear Simultaneous Equations | How to Determine if a Linear Equation is Solvable or Not?"},"content":{"rendered":"

Learn the conditions for the solvability of a system of linear equations in two variables in the following sections. If the simultaneous equations have no solution, then they are called inconsistent and if they have a solution, then they are called consistent. To help you understand the fundamentals of the concept we have given the Solved Examples with Answers explaining everything in detail.<\/p>\n

Conditions for Solvability of Simultaneous Linear Equations in Two Variables<\/h2>\n

Let the pair of linear equations in two variables are<\/p>\n

a\u2081x + b\u2081y + c\u2081 = 0 ——- (1)<\/p>\n

a\u2082x + b\u2082y + c\u2082= 0 ——- (2)<\/p>\n

By using the cross-multiplication method, we get<\/p>\n

x\/(b\u2081c\u2082 – b\u2082c\u2081) = y\/(a\u2082c\u2081 – a\u2081c\u2082) = 1\/(a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

So, x = (b\u2081c\u2082 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

y = (a\u2082c\u2081 – a\u2081c\u2082) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

Now see when the solvability of simultaneous linear equations in two variables (i), (ii) are solavble.<\/p>\n

(I) If (a\u2081b\u2082 – a\u2082b\u2081) \u2260 0 for any values of (b\u2081c\u2082 – b\u2082c\u2081), (a\u2082c\u2081 – a\u2081c\u2082) we will get the unique solutions for the x and y variables.<\/p>\n

Example:<\/strong><\/p>\n

x – 2y – 8 = 0 ——- (1)<\/p>\n

x + y – 5 = 0 ——— (2)<\/p>\n

Here, a\u2081 = 1, b\u2081 = -2, c\u2081 = -8, a\u2082 = 1, b\u2082 = 1, c\u2082 = -5<\/p>\n

Substitute these values in (a\u2081b\u2082 – a\u2082b\u2081), we get<\/p>\n

(a\u2081b\u2082 – a\u2082b\u2081) = (1 x 1 – 1 x -2) = 1 + 2 = 3 \u2260 0<\/p>\n

So, x = (b\u2081c\u2082 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

= (-2 x -5 – 1 x -8) \/ 3 = (-10 + 8)\/3 = 18\/3 = 6<\/p>\n

y = (a\u2082c\u2081 – a\u2081c\u2082) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

= (1 x -8 – 1 x -5)\/3 = (-8 + 5)\/3 = -3\/3 = -1<\/p>\n

Therefore, when (a\u2081b\u2082 – a\u2082b\u2081) \u2260 0, then the system of linear equations are always consistent.<\/p>\n

(II) If (a\u2081b\u2082 – a\u2082b\u2081) = 0 and any one of (b\u2081c\u2082 – b\u2082c\u2081) and (a\u2082c\u2081 – a\u2081c\u2082) is zero, then<\/p>\n

Let us take a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 = k, where k \u2260 0<\/p>\n

Then, a\u2081 = ka\u2082, b\u2081 = kb\u2082, c\u2081 = kc\u2082<\/p>\n

And system of linear equations are changed to<\/p>\n

ka\u2082x + kb\u2082y + kc\u2082 = 0<\/p>\n

a\u2082x + b\u2082y + c\u2082 = 0<\/p>\n

From this equation, we can write<\/p>\n

x = (-b\u2082y – c\u2082) \/ a\u2082<\/p>\n

This indicates for each value of y, there is a definite value of x or there are infinite number of solutions of the simultaneous equations in this case.<\/p>\n

Example:<\/strong><\/p>\n

x + 2y + 5 = 0<\/p>\n

3x + 6y + 15 = 0<\/p>\n

Here, a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 = 1\/3<\/p>\n

Actually, we get the second equation by multiplying the first equation by 3. Express one equation x interms of y<\/p>\n

x = -2y -5<\/p>\n

Some of the solutions are<\/p>\n\n\n\n\n\n
y<\/th>\n-1<\/th>\n0<\/th>\n1<\/th>\n2<\/th>\n. . . . . . .<\/th>\n<\/tr>\n<\/thead>\n
x<\/td>\n-3<\/td>\n-5<\/td>\n-7<\/td>\n-9<\/td>\n. . . . .<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(III) If (a\u2081b\u2082 – a\u2082b\u2081) = 0 and any one from (b\u2081c\u2082 – b\u2082c\u2081) and (a\u2082c\u2081 – a\u2081c\u2082) is non zero, (then other one is also non-zero)<\/p>\n

Let us take k = a\u2081\/a\u2082 = b\u2081\/b\u2082 \u2260 c\u2081\/c\u2082 = k<\/p>\n

So, a\u2081 = ka\u2082, b\u2081 = kb\u2082<\/p>\n

In this case, the chnaged linear equations are<\/p>\n

ka\u2082x + kb\u2082y + c\u2082 = 0<\/p>\n

a\u2082x + b\u2082y + c\u2082= 0<\/p>\n

These equations do not have solution for x and y.<\/p>\n

So the equations are inconsistent.<\/p>\n

While drawing graphs, we can see that the linear equations in two variables always represent a straight line and the equations of the above form represent two parallel straight lines. This is why these lines do not intersect each other and not have a common point.<\/p>\n

Example:<\/strong><\/p>\n

7x + y + 3 = 0<\/p>\n

14x + 2y – 1 = 0<\/p>\n

Here, a\u2081 = 7, b\u2081 = 1, c\u2081 = 3, a\u2082 = 14, b\u2082 = 2, c\u2082 = -1<\/p>\n

And, a\u2081\/a\u2082 = b\u2081\/b\u2082 \u2260 c\u2081\/c\u2082<\/p>\n

So, the given system of linear equations is inconsistent.<\/p>\n

From the above discussions, we can write that<\/p>\n

a\u2081x + b\u2081y + c\u2081 = 0, a\u2082x + b\u2082y + c\u2082= 0 will be<\/p>\n

(i) Consistent if a\u2081\/a\u2082 \u2260 b\u2081\/b\u2082, we will get a unique solution<\/p>\n

(ii) Inconsistent, there will be no solution if a\u2081\/a\u2082 = b\u2081\/b\u2082 \u2260 c\u2081\/c\u2082 where c\u2081 \u2260 0, c\u2082 \u2260 0<\/p>\n

(iii) Consistent and have infinite solutions if a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 where c\u2081 \u2260 0, c\u2082 \u2260 0<\/p>\n

Solvability of Linear Equations Examples<\/h3>\n

Example 1.<\/strong><\/p>\n

Check whether the linear equations x + 2y + 4 = 0, 3x – 5y + 1 = 0 are consistent or inconsistent?<\/p>\n

Solution:<\/strong><\/p>\n

Given system of linear equations are<\/p>\n

x + 2y + 4 = 0<\/p>\n

3x – 5y + 1 = 0<\/p>\n

Here, a\u2081 = 1, b\u2081 = 2, c\u2081 = 4, a\u2082 = 3, b\u2082 = -5, c\u2082 = 1<\/p>\n

(a\u2081b\u2082 – a\u2082b\u2081) = (1 x -5 – 3 x 2) = (-5 – 6) = -11<\/p>\n

x = (b\u2081c\u2082 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

= (2 x 1 – (-5) x 4) \/ -11 = (2 + 20)\/-11<\/p>\n

= -22\/11 = -2<\/p>\n

y = (a\u2082c\u2081 – a\u2081c\u2082) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

= (3 x 4 – 1 x 1) \/ -11 = (12 – 1)\/-11 = -11\/11<\/p>\n

= -1<\/p>\n

Here, (a\u2081b\u2082 – a\u2082b\u2081) \u2260 0<\/p>\n

Therefore, the given simultaneous linear equations are consistent and have a unique solution.<\/p>\n

Example 2.<\/strong><\/p>\n

Check whether the linear simultaneous equations 3x – y – 2 = 0, 6x – 2y – 4 = 0 are consistent or inconsistent?<\/p>\n

Solution:<\/strong><\/p>\n

Given Linear Simultaneous Equations are<\/p>\n

3x – y – 2 = 0, 6x – 2y – 4 = 0<\/p>\n

Here, a\u2081 = 3, b\u2081 = -1, c\u2081 = -2, a\u2082 = 6, b\u2082 = -2, c\u2082 = -4<\/p>\n

(a\u2081b\u2082 – a\u2082b\u2081) = (3 x -2 – 6 x -1) = (-6 + 6) = 0<\/p>\n

and (b\u2081c\u2082 – b\u2082c\u2081) = (-1 x -4 – (-2) x -2) = (4 – 4) = 0<\/p>\n

(a\u2082c\u2081 – a\u2081c\u2082) = (6 x -2 – 3 x (-4)) = (-12 + 12) = 0<\/p>\n

a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 = 1\/2<\/p>\n

Actually, we get the second equation by multiplying the first equation by 2. Express one equation x in terms of y<\/p>\n

x = (y + 2) \/ 3<\/p>\n

Some of the solutions are<\/p>\n\n\n\n\n\n
y<\/th>\n-1<\/th>\n0<\/th>\n1<\/th>\n2<\/th>\n. . .<\/th>\n<\/tr>\n<\/thead>\n
x<\/td>\n1\/3<\/td>\n2\/3<\/td>\n1<\/td>\n4\/3<\/td>\n. . .<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Therefore, the given linear equations are consistent and have an infinite number of solutions.<\/p>\n

Example 3.<\/strong><\/p>\n

State whether the linear equations -2x + 3y – 2 = 0, 2x – 3y – 5 = 0 are consistent or not?<\/p>\n

Solution:<\/strong><\/p>\n

Given linear equations are -2x + 3y – 2 = 0, 2x – 3y – 5 = 0<\/p>\n

Here, a\u2081 = -2, b\u2081 = 3, c\u2081 = -2, a\u2082 = 2, b\u2082 = -3, c\u2082 = -5<\/p>\n

(a\u2081b\u2082 – a\u2082b\u2081) = (-2 x -3 – 2 x 3) = (6 – 6) = 0<\/p>\n

(b\u2081c\u2082 – b\u2082c\u2081) = (3 x -5 – (-3) x (-2)) = (-15 – 6) = -21<\/p>\n

a\u2081\/a\u2082 = b\u2081\/b\u2082 = -1<\/p>\n

The equations are inconsistent and has no solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Learn the conditions for the solvability of a system of linear equations in two variables in the following sections. If the simultaneous equations have no solution, then they are called inconsistent and if they have a solution, then they are called consistent. To help you understand the fundamentals of the concept we have given the … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35227"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=35227"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35227\/revisions"}],"predecessor-version":[{"id":138473,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35227\/revisions\/138473"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=35227"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=35227"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=35227"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}