Algebra - Solved Examples

1.    If (x + 2)(x - a) = px2$x^2$ + qx + 8, what are the values of the constants a, p and q?
Solution: (x + 2)(x - a) = px2$p{x}^2$ + qx + 8
⇒x2$\Rightarrow x^2$ + 2x - ax - 2a = px2$p{x}^2$ + qx + 8
⇒x2$\Rightarrow x^2$ + (2 - a) x - 2a = px2$p{x}^2$ + qx + 8
Equating the coefficients of x2$x^2$ , x and constant terms on both sides.
p = 1; q = 2 - a
-2a = 8
Solving, we get a = -4, p = 1, q = 6

2.    (x +1 )(2x - 2)(3x + 3) =  ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$ What are the values of a, b, c and d?
Solution: (x + 1)(2x - 2)(3x + 3) = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
6(x + 1)(x - 1)(x + 1) = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
6(x2$x^2$ - 1)(x + 1) = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
⇒6(x3+x2−x−1)$\Rightarrow 6\left(x^3+x^2-x-1\right)$ = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
⇒6x3+6x2−6x−6$\Rightarrow 6x^3+6x^2-6x-6$ = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
    Equating the coefficients of like learns on both the sides, a = 6, b = 6, c = -6 and d = -6.

3.    Find for what value(s) of k would there be an unique solution for the given set of questions.
    2x - 3y = 1 and kx + 5y = 7
Solution: If two equations ax + by = M and cx + dy = N have a unique solution, then ac≠bd${\displaystyle \frac{a}{c}\ne {\displaystyle \frac{b}{d}}}$
 So in the above problem, k ≠−103$\ne {\displaystyle \frac{-10}{3}}$

4.    Find the value(s) of k for which there is no solution for the given set of equations.
    2x - ky = -3 and 3x + 2y = 1
Solution: 3x + 2y = 1
If there are no set of solutions, then 2−k=32${\displaystyle \frac{2}{-k}={\displaystyle \frac{3}{2}}}$ or k = −43$-{\displaystyle \frac{4}{3}}$
    For k = −43$-{\displaystyle \frac{4}{3}}$, the two lines in the coordinate planes are parallel to each other.

5.    Find the value of k for which there are infinite solutions for the given set of equations.
    5x + 2y = k and 10x + 4y = 3
Solution:  For the two sets of equation to have infinite solutions, we have 510=24=k3${\displaystyle \frac{5}{10}={\displaystyle \frac{2}{4}={\displaystyle \frac{k}{3}}}}$
    . Hence, k = 32${\displaystyle \frac{3}{2}}$

6.    What is the solution of the following simultaneous equations?
    x + y + z = 6, x + 2y + 3z = 14 and x + 3y + z = 10
Solution:  x + y + z = 6        .... (i)
x + 2y + 3z = 14    .... (ii)
x + 3y + z = 10
From (i), we get z = 6 - x - y
Substitute it in (ii) and (iii)
We have from (ii)
x + 2y + 18 - 3x - 3y = 14 or 2x + y = 4
Similarily, from (iii)
x + 3y + 6 - x - y = 10 or 2y = 4 or y = 2
On solving, we get y = 2, x = 1, z = 3.