Question

I. $8x+6y=(6400{)}^{\frac{1}{2}}$
II. $18x-15y=(4356{)}^{\frac{1}{2}}$

• A. $x>y$
• B. $x\ge y$
• C. $x<y$
• D. $x\le y$
• E. $x=y$or relationship cannot be established

Answer 1

The correct option is A $x>y$

Equation I
$8x+6y=(6400{)}^{\frac{1}{2}}$
$4x+3y=40$ ...(i)
Equation II
$18x-15y=(4536{)}^{\frac{1}{2}}$
$6x-5y=22$ ...(ii)
On solving Eqs. (i) and (ii), we get
4x + 3y = 40
6x - 5y = 22
On multiplying Eq. (i) by 6 and Eq. (ii) by 4,
$24x+18y=24024x-20y=88-+-\underset{–}{}38y=152$
$\therefore y=\frac{152}{38}=4$
On putting the value of y in Eq. (ii), we get
6x - 5(4) = 22
$\Rightarrow 6x=22+20\Rightarrow 6x=42\therefore x=7$
Hence, x > y