Question

The linear equation $5x+12y=48$ cuts the x-axis at

• A. $(0,4)$
• B. $(4,0)$
• C. $(0,\frac{48}{5})$
• D. $(\frac{48}{5},0)$

Answer 1

The correct option is D $(\frac{48}{5},0)$

Any point that intersects the x-axis is in the form of (k, 0), where k is any real number.
Substituting y = 0 in $5x+12y=48$, we get
$5x+12\left(0\right)=48$
$\Rightarrow 5x=48$
$\Rightarrow x=\frac{48}{5}$
Therefore, the given linear equation intersects the x-axis at $(\frac{48}{5},0)$
Hence, the correct answer is option (d).