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$\esone{p x + 3 y = [p - 3]} $
\left\{\begin{matrix}p=-\frac{3\left[y+1\right]}{x-1}\text{, }&x\neq 1\\p\in \mathrm{R}\text{, }&y=-1\text{ and }x=1\end{matrix}\right.
\left\{\begin{matrix}x=-\frac{3y-p+3}{p}\text{, }&p\neq 0\\x\in \mathrm{R}\text{, }&p=0\text{ and }y=-1\end{matrix}\right.
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px+3y-p=-3
Subtract p from both sides.
px-p=-3-3y
Subtract 3y from both sides.
\left[x-1\right]p=-3-3y
Combine all terms containing p.
\left[x-1\right]p=-3y-3
The equation is in standard form.
\frac{\left[x-1\right]p}{x-1}=\frac{-3y-3}{x-1}
Divide both sides by x-1.
p=\frac{-3y-3}{x-1}
Dividing by x-1 undoes the multiplication by x-1.
p=-\frac{3\left[y+1\right]}{x-1}
Divide -3-3y by x-1.
px=p-3-3y
Subtract 3y from both sides.
px=-3y+p-3
The equation is in standard form.
\frac{px}{p}=\frac{-3y+p-3}{p}
Divide both sides by p.
x=\frac{-3y+p-3}{p}
Dividing by p undoes the multiplication by p.
Examples
Question 12 - CBSE Class 10 Sample Paper for 2019 Boards - Solutions of Sample Papers for Class 10 Boards
Last updated at Sept. 24, 2021 by
Question 12
For what value of p will the following pair of linear equations have infinitely many solutions
[p – 3]x + 3y = p
px + py = 12
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Transcript
Question 12 For what value of p will the following pair of linear equations have infinitely many solutions [p – 3]x + 3y = p px + py = 12 Given equations [p – 3]x + 3y = p px + py = 12 [p – 3]x + 3y = p [p – 3]x + 3y – p = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = [p – 3] , b1 = 3 , c1 = –p px + py = 12 px + py – 12 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = p , b2 = p , c2 = –12 Given that Equation has infinite number of solutions ∴ 𝑎1/𝑎2 = 𝑏1/𝑏2 = 𝑐1/𝑐2 Putting in values [[𝑝 − 3]]/𝑝 = 3/𝑝 = [−𝑝]/[−12] [[𝑝 − 3]]/𝑝 = 3/𝑝 = 𝑝/12 1/2 marks Solving [[𝒑 − 𝟑]]/𝒑 = 𝟑/𝒑 p[p – 3] = 3p p2 – 3p = 3p p2 – 3p – 3p = 0 p2 – 6p = 0 p[p – 6] = 0 So, p = 0, 6 1/2 marks Solving 𝟑/𝒑 = 𝒑/𝟏𝟐 3 × 12 = p2 36 = p2 p2 = 36 p = ± √36 p = ± 6 So, p = 6, –6 1/2 marks Since p = 6 satisfies both equations. Hence, p = 6 is the answer 1/2 marks