Question

The point at which two lines 3x + 4y = 30 and 7x – 2y = 35 cut each other is

• A. $\frac{100}{17},\frac{105}{34}$
• B. $\frac{100}{17},\frac{105}{17}$
• C. $\frac{105}{34},\frac{100}{17}$
• D. $\frac{100}{34},\frac{105}{17}$

Answer 1

The correct option is A $\frac{100}{17},\frac{105}{34}$

Given: 3x + 4y = 30 ..…(i)
and, 7x – 2y = 35 ..…(ii)
Multiplying (ii) by 2 and adding to (i) we get,
(3x + 4y) + 2(7x - 2y) = 30+$2\times 35)\Rightarrow 3x+4y+14x-4y=30+70\Rightarrow 17x=100\Rightarrow x=\frac{100}{17}$
Now, substituting, $x=\frac{100}{17}$ in (i) we get,
$\frac{100}{17}+4y=30\Rightarrow \frac{300}{17}+4y=30\Rightarrow 4y=30-\frac{300}{17}\Rightarrow 4y=\frac{210}{17}\Rightarrow y=\frac{105}{34}$
Thus, $\frac{100}{17},\frac{105}{34}$ is the point at which the given lines cut each other.
Hence, the correct answer is option (1).