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How to solve the system of linear equations in a easy way? How to solve the system of linear equations in a easy way?
How to solve the system of linear equations in a easy way?
Dear Suresh Solving systems of equations graphically is one of the easiest ways to solve systems of simple equations Another way to solve systems of equations is by substitution. In this method, you solve on equation for one variable, then you substitute that solution in the other equation, and solve. Example: 1. Problem: Solve the following system: x + y = 11 3x - y = 5 Solution: Solve the first equation for y (you could solve for x - it doesn't matter). y = 11 - x Now, substitute 11 - x for y in the second equation. This gives the equation one variable, which earlier algebra work has taught you how to do. 3x - (11 - x) = 5 3x - 11 + x = 5 4x = 16 x = 4 Now, substitute 4 for x in either equation and solve for y. (We use the first equation below.) 4 + y = 11 y = 7Solve the following system: x + y + z = 4 x - 2y - z = 1 2x - y - 2z = -1 Solution: Start out by multiplying the first equation by -1 and add it to the second equation to eliminate x from the second equation. -x - y - z = -4 x - 2y - z = 1 ---------------- -3y - 2z = -3 Now eliminate x from the third equation by multiplying the first equation by -2 and add it to the third equation. -2x - 2y - 2z = -8 2x - y - 2z = -1 ------------------ -3y - 4z = -9 Next, eliminate y from the third equation by multiplying the second equation by -1 and adding it to the third equation. 3y + 2z = 3 -3y - 4z = -9 -------------- -2z = -6 Solve the third equation for z. -2z = -6 z = 3 Substitute 3 for z in the second equation and solve for y. -3y - 2z = -3 -3y - 2(3) = -3 -3y - 6 = -3 -3y = 3 y = -1 Lastly, substitute -1 for y and 3 for z in the first equation and solve for x. x + (-1) + 3 = 4 x + 2 = 4 x = 2 All the best. AKASH GOYAL AskiitiansExpert-IITD Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Dear Suresh
Solving systems of equations graphically is one of the easiest ways to solve systems of simple equations
Another way to solve systems of equations is by substitution. In this method, you solve on equation for one variable, then you substitute that solution in the other equation, and solve. Example:
1. Problem: Solve the following system: x + y = 11 3x - y = 5 Solution: Solve the first equation for y (you could solve for x - it doesn't matter). y = 11 - x Now, substitute 11 - x for y in the second equation. This gives the equation one variable, which earlier algebra work has taught you how to do. 3x - (11 - x) = 5 3x - 11 + x = 5 4x = 16 x = 4 Now, substitute 4 for x in either equation and solve for y. (We use the first equation below.) 4 + y = 11 y = 7Solve the following system: x + y + z = 4 x - 2y - z = 1 2x - y - 2z = -1 Solution: Start out by multiplying the first equation by -1 and add it to the second equation to eliminate x from the second equation. -x - y - z = -4 x - 2y - z = 1 ---------------- -3y - 2z = -3 Now eliminate x from the third equation by multiplying the first equation by -2 and add it to the third equation. -2x - 2y - 2z = -8 2x - y - 2z = -1 ------------------ -3y - 4z = -9 Next, eliminate y from the third equation by multiplying the second equation by -1 and adding it to the third equation. 3y + 2z = 3 -3y - 4z = -9 -------------- -2z = -6 Solve the third equation for z. -2z = -6 z = 3 Substitute 3 for z in the second equation and solve for y. -3y - 2z = -3 -3y - 2(3) = -3 -3y - 6 = -3 -3y = 3 y = -1 Lastly, substitute -1 for y and 3 for z in the first equation and solve for x. x + (-1) + 3 = 4 x + 2 = 4 x = 2 All the best. AKASH GOYAL AskiitiansExpert-IITD Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Solve the following system: x + y + z = 4 x - 2y - z = 1 2x - y - 2z = -1 Solution: Start out by multiplying the first equation by -1 and add it to the second equation to eliminate x from the second equation. -x - y - z = -4 x - 2y - z = 1 ---------------- -3y - 2z = -3 Now eliminate x from the third equation by multiplying the first equation by -2 and add it to the third equation. -2x - 2y - 2z = -8 2x - y - 2z = -1 ------------------ -3y - 4z = -9 Next, eliminate y from the third equation by multiplying the second equation by -1 and adding it to the third equation. 3y + 2z = 3 -3y - 4z = -9 -------------- -2z = -6 Solve the third equation for z. -2z = -6 z = 3 Substitute 3 for z in the second equation and solve for y. -3y - 2z = -3 -3y - 2(3) = -3 -3y - 6 = -3 -3y = 3 y = -1 Lastly, substitute -1 for y and 3 for z in the first equation and solve for x. x + (-1) + 3 = 4 x + 2 = 4 x = 2
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
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