RD Sharma Solutions Class 12 Mathematics Chapter 7 VSA

# RD Sharma Solutions Class 12 Mathematics Chapter 7 VSA

Edited By Kuldeep Maurya | Updated on Jan 20, 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

$\rightarrow x=\frac{2}{3}, y=-2$
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

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

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

$\rightarrow 2$
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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## RD Sharma Chapter wise Solutions

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1. Where can I download the RD Sharma class 12 VSA solution?

You can download the RD Sharma class 12 VSA from the Career360 website for free, as no penny will be charged for the same.

2. Does the RD Sharma class 12 chapter 7 VSA have the latest syllabus?

Yes, it has the latest version, as the syllabus is regularly updated to correspond with the syllabus of NCERT. So students can also refer to this solution for the preparation of public exams.

3. Can I use RD Sharma class 12 VSA for solving homework?

Yes, you can use the solutions in the RD Sharma VSA book to solve the homework assigned to you. As teachers refer to these solutions for assigning homework, it helps you in solving and saves time.

4. Is Simultaneous linear equation an easy chapter to solve?

Well, everything becomes easier if you practice the sums with the RD Sharma class 12 chapter 7 VSA book. It makes the process quicker.

5. Can the solutions in the RD Sharma books be trusted?

Everything present in the RD Sharma class 12 exercise VSA  is verified by experts in the educational sector. As experts in the academic field prepare the solutions, the questions provided are correct and verified.

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