Question

A system of linear equations in two variables, p and q, is given as (n + 1)p+ (n + 2)q = 8, p - (n + 1)q + (n + 2) = 0, p + q = 3. Which of the following would be one of the values of 'n' for which the given system of linear equations is consistent?​

Answer 1

Answer:

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Answer 2

Tip:

• If $|A| \neq 0$ , then system of linear equation is consistent and has unique solution.
• If $|A|=0$ and $[adj (A)]B=0$ i.e., null matrix, then the system of linear equation is consistent and has infinitely many solutions.

Explanation:

• We have a system of linear equation in two variables, $p$ and $q$ as $(n+1)p+(n+2)q=8\\p-(n+2)q=0\\p+q=3$

• We have to find the value of $n$ if the system of linear equation is consistent.
• We will solve this question by criteria of consistency.

Steps:

Step 1 of 2:

The given system of linear equations is

$(n+1)p+(n+2)q=8\\p-(n+2)q=0\\p+q=3$

This system is consistent, if $\left|\begin{array}{ccc}n+1&n+2&8\\1&-n-1&-n-2\\1&1&3\end{array}\right| =0$[apply $R_3$↔$R_1$]

⇒ $\left|\begin{array}{ccc}1&1&3\\1&-n-1&-n-2\\n+1&n+2&8\end{array}\right|$ [apply $R_2\rightarrow R_2-R_1\\R_3\rightarrow R_3-(n+1)R_1$]

⇒  $\left|\begin{array}{ccc}1&1&3\\0&-n-2&-n-5\\0&1&-3n+5\end{array}\right|$  

Step 2 of 2:

⇒ $(-n-2)(-3n+5)-1(-n-5)=0$

⇒ $3n^2-5n+6n-10+n+5=0$

⇒ $3n^2+2n-5=0$

⇒ $3n^2+5n-3n-5=0$

⇒ $(n-1)(3n+5)=0$

⇒ $n=1,n=-\frac{5}{3}$

Final Answer:

The values of $n$ for which the given system of linear equations is consistent is  $n=1,n=-\frac{5}{3}$