Name: Ratio of Greater and Lesser Inequalities
Uploaded: 2019-07-19
Duration: 5 min 30 s
Description: In this video, learn a basic introduction to solving rational inequalities using a sign chart on a number line and expressing the solution using interval notation.

Question

If

\frac{2 y + 2 z - x}{a} = \frac{2 z + 2 x - y}{b} = \frac{2 x + 2 y - z}{c}

, show that

\frac{x}{2 b + 2 c - a} = \frac{y}{2 c + 2 a - b} = \frac{z}{2 a + 2 b - c} .

Accepted Answer

Given,

$\frac{2 y + 2 z - x}{a} = \frac{2 z + 2 x - y}{b} = \frac{2 x + 2 y - z}{c}$

So,

$\frac{2 y + 2 z - x}{a} = \frac{2 z + 2 x - y}{b}$

$\Rightarrow$ $2 b y + 2 b z - b x = 2 a z + 2 a x - a y$

$\Rightarrow x (b + 2 a) - y (2 b + a) + z (2 a - 2 b) = 0 \dots (i)$

Again,

$\frac{2 z + 2 x - y}{b} = \frac{2 x + 2 y - z}{c}$

$\Rightarrow 2 b x + 2 b y - b z = 2 c z + 2 c x - c y .$

$\Rightarrow$ $x (2 b - 2 c) + y (2 b + c) - z (b + 2 c) = 0 \dots (ii)$

Applying cross multiplication on $(i)$ and $(ii)$ we get,

$\Rightarrow \frac{x}{[(b + 2 c) (a + 2 b) - 2 (2 b + c) (a - b)]}$ $= \frac{y}{[2 (a - b) (2 b - 2 c) + (2 a + b) (b + 2 c)]}$ $= \frac{z}{[(2 b + c) (2 a + b) + (a + 2 b) (2 b - 2 c)]}$

$\Rightarrow \frac{x}{a b + 2 a c + 2 b^{2} + 4 b c - 4 a b + 4 b^{2} - 2 c a + 2 b c} = \frac{y}{4 a b - 4 a c - 4 b^{2} + 4 b c + 2 a b + 4 a c + b^{2} + 2 b c} = \frac{z}{4 a b + 2 a c + 2 b^{2} + b c + 2 a b + 4 b^{2} - 2 a c - 4 b c}$

$\Rightarrow \frac{x}{6 b c + 6 b^{2} - 3 a b} = \frac{y}{6 a b - 3 b^{2} + 6 b c} = \frac{z}{6 a b + 6 b^{2} - 3 b c}$

$\Rightarrow \frac{x}{2 c + 2 b - a} = \frac{y}{2 a + 2 c - b} = \frac{z}{2 b + 2 a - c}$

Hence, proved.

If $\frac{2 y + 2 z - x}{a} = \frac{2 z + 2 x - y}{b} = \frac{2 x + 2 y - z}{c}$ , show that $\frac{x}{2 b + 2 c - a} = \frac{y}{2 c + 2 a - b} = \frac{z}{2 a + 2 b - c} .$

Important Questions on Ratio

Learn and Prepare for any exam you want !

If 2y+2z-xa=2z+2x-yb=2x+2y-zc, show that x2b+2c-a=y2c+2a-b=z2a+2b-c.

Important Questions on Ratio

Learn and Prepare for any exam you want !

If $\frac{2 y + 2 z - x}{a} = \frac{2 z + 2 x - y}{b} = \frac{2 x + 2 y - z}{c}$ , show that $\frac{x}{2 b + 2 c - a} = \frac{y}{2 c + 2 a - b} = \frac{z}{2 a + 2 b - c} .$