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Is Simultaneous linear equation an easy chapter to solve?

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RD Sharma Solutions Class 12 Mathematics Chapter 7 VSA

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RD Sharma Solutions Class 12 Mathematics Chapter 7 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 7 VSA

Updated on 20 Jan 2022, 03:15 PM IST

The Class 12 RD Sharma chapter & Exercise VSA solution books are used by every student who is preparing for their public examinations. So, when it comes to mathematics, most of the students struggle to find solutions while doing their homework. Even though chapters like simultaneous linear equations are not so challenging, students cannot find the answers for the problems related to the branch. Therefore, it is highly recommended that the students use the RD Sharma class 12 solution of simultaneous linear equation exercise VSA for their practical help.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 7 VSA Solution of Simultaneous Linear Equation - Other Exercise

Solution of Simultaneous Linear Equation Exercise Fill in the blank Question 1

Answer $\rightarrow x=1, y=-1, z=0$
Hint $\rightarrow I A=A$
Given
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]$
Explanation
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]$
$\left[\begin{array}{c} x \times 1+0 \times y+0 \times z \\ 0 \times x+1 \times y+0 \times z \\ 0 \times x+0 \times y+1 \times z \end{array}\right]=\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]$
$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ -1 \\ 0 \end{array}\right]$
Comparing both sides we get $x=1,y=-1,z=0$

Solution of Simultaneous Linear Exercise Very short answer Question 2

Answer $\rightarrow x=1, y=0, z=-1$
Given
$\rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$
Explanation
$\rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$
$\left[\begin{array}{l} x \times 1+0 \times y+0 \times z \\ 0 \times x-1 \times y+0 \times z \\ 0 \times x+0 \times y-1 \times z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$
$\left[\begin{array}{c} x \\ -y \\ -z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$
Comparing both sides we get $x=1,y=0,z=-1$

Solution of Simultaneous Linear Equation Exercise Very short answer Question 3

Answer $\rightarrow x=1,y=0,z=1$
Given
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ -1 \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$
Explanation
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ -1 \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$
$\begin{aligned} &{\left[\begin{array}{l} 1 \times x+0 \times-1+0 \times z \\ 0 \times x+y \times-1+0 \times z \\ 0 \times x+0 \times-1+1 \times z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]} \\\\ &{\left[\begin{array}{c} x \\ -y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]} \end{aligned}$
Comparing both sides we get $x=1,y=0,z=1$

Solution of Simultaneous Linear Equation Exercise Very short answer Question 4

Answer
$\rightarrow x=\frac{2}{3}, y=-2$
Given
$\rightarrow\left[\begin{array}{cc} 3 & -4 \\ 9 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right]$
Explanation
$\begin{gathered} \rightarrow\left[\begin{array}{cc} 3 & -4 \\ 9 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right] \\\\ \quad\left[\begin{array}{c} 3 \times x-4 \times y \\ 9 \times x+2 \times y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right] \\\\ {\left[\begin{array}{c} 3 x-4 y \\ 9 x+2 y \end{array}\right]=\left[\begin{array}{c} 10 \\ 2 \end{array}\right]} \end{gathered}$
Comparing both sides we get
$\begin{aligned} &\\ &3 x-4 y=10-----(1)\\ &9 x+2 y=2------(2)\\ &\text { } \end{aligned}$
Multiplying eqn. 1 by (3) and then subtracting eqn. 2 from it
$\begin{aligned} &9 x-12 y=30 \\ &9 x+2 y\: \: =2 \\ &-14 y\: \: \: \: =28 \\ &\: \: \: \: \; \; \; \; \; \; \; \; y=-2 \end{aligned}$
Put this value in (1)
$3 x-4(-2)=10 \quad \quad \quad[3 x-4 y=10]$
$\begin{aligned} &3 x=10-8 \\ &x=2 / 3 \end{aligned}$
Hense
$x=2 / 3, y=-2$

Solution of Simultaneous Linear Equation Exercise Very short answer Question 5

Answer
$\rightarrow x=2, y=3, z=-1$

Given
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$
Explanation
$\rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$
$\begin{aligned} &{\left[\begin{array}{l} 1 \times x+0 \times y+0 \times z \\ 0 \times x+0 \times y+1 \times z \\ 0 \times x+1 \times y+0 \times z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]} \\\\ &{\left[\begin{array}{l} x \\ z \\ y \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]} \end{aligned}$
Comparing both sides we get $x=2, y=3, z=-1$

Solution of Simultaneous Linear Equation Exercise Very short answer Question 6

Answer
$\rightarrow 2$
Given
$\rightarrow A=\left[\begin{array}{ll} 2 & 4 \\ 4 & 3 \end{array}\right], \quad X=\left[\begin{array}{l} n \\ 1 \end{array}\right], B=\left[\begin{array}{c} 8 \\ 11 \end{array}\right], A X=B$
Explanation
$\rightarrow AX=B$
$\left[\begin{array}{ll} 2 & 4 \\ 4 & 3 \end{array}\right]\left[\begin{array}{l} n \\ 1 \end{array}\right]=\left[\begin{array}{c} 8 \\ 11 \end{array}\right]$
$\begin{aligned} &{\left[\begin{array}{c} 2 \times n+4 \times 1 \\ 4 \times n+3 \times 1 \end{array}\right]=\left[\begin{array}{c} 8 \\ 11 \end{array}\right]} \\\\ &{\left[\begin{array}{l} 2 n+4 \\ 4 n+3 \end{array}\right]=\left[\begin{array}{c} 8 \\ 11 \end{array}\right]} \end{aligned}$
Comparing both sides we get
$\begin{aligned} &2 n+4=8 \\ &2 n=8-4 \\ &2 n=4 \end{aligned}$
$\begin{aligned} &n=\frac{4}{2} \\ &n=2 \end{aligned}$
$\begin{aligned} &4 n+3=11 \\ &4 n=11-3 \\ &4 n=8 \end{aligned}$
$\begin{aligned} &n=8 / 4 \\ &n=2 \end{aligned}$
Hence $n=2$

The RD Sharma class 12th exercise VSA of Simultaneous linear equation is quite challenging for most students, and the absence of appropriate resources makes things worse for them. Therefore, to save the students from such a situation, the RD Sharma class 12th exercise VSA solution book is recommended as the questions are handpicked by experts in the field of mathematics, providing helpful tips to the students that they might not be getting in school as well. There are only six questions in the RD Sharma class 12 solutions chapter 7 exercise VSA, and few concepts are covered under the exercise, that are,

Methods to solve linear equations
Solutions of the equation lying on a straight line
Solving linear equations in two variables
Solving linear equations in three variables

Therefore, the concepts covered in the RD Sharma class 12th exercise VSA are enough for any student to practice and build up their performance in the maths subject which helps them to score high in their board exam.

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The most significant advantage is that the students will not have to pay a single penny to own this book. They can download the free E-books and PDF of the RD Sharma class 12th exercise VSA for free from the Career360 website. Not only students but teachers can also refer to the Career360 website to download any book of any subject. There is an abundance of books referred to students that can be available at this site. Using the RD Sharma class 12th exercise VSA, you will face no difficulty finding the answers for the concept in a Simultaneous linear equation.