Question
Find four solutions for the following equation $:12x + 5y = 0$
✓
Answer
$12 x+5 y=0$
$\Rightarrow 5 y=-12 x$
$\Rightarrow y=\frac{-12}{5} x$
Put $x=0$, then $y=\frac{-12}{5}(0)=0$
Put $x=5$, then $y=\frac{-12}{5}(5)=-12$
Put $x=10$, then $y=\frac{-12}{5}(10)=-24$
Put $x=15$, then $y=\frac{-12}{5}(15)=-36$
$\therefore(0,0),(5,-12),(10,-24)$ and $(15,-36)$ are the four solutions of the equation $12 x+5 y=0$
$\Rightarrow 5 y=-12 x$
$\Rightarrow y=\frac{-12}{5} x$
Put $x=0$, then $y=\frac{-12}{5}(0)=0$
Put $x=5$, then $y=\frac{-12}{5}(5)=-12$
Put $x=10$, then $y=\frac{-12}{5}(10)=-24$
Put $x=15$, then $y=\frac{-12}{5}(15)=-36$
$\therefore(0,0),(5,-12),(10,-24)$ and $(15,-36)$ are the four solutions of the equation $12 x+5 y=0$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
There are two cones. The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.
If $O$ is the centre of the circle, find the value of $x$ in the following figure:
→Show that a diagonal divides a parallelogram into two triangles of equal area.
If $x+y+z=0$, show that $x^3+y^3+z^3=3xyz$.
→Determine the point on the graph of the linear equation $2x + 5y = 19$ whose ordinate is $1\frac{1}{2}$ times its abscissa.
→Prove that: $\frac{1}{1+\text{x}^{\text{a}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}}=1$
→Prove that two different circles cannot intersect each other at more than two points.
Construct a histogram for the following data:
→| Monthly School fee (in ₹): | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $150-180$ | $180-210$ | $210-240$ |
| No of schools | $5$ | $12$ | $14$ | $18$ | $10$ | $9$ | $4$ |
Find the following products:
$(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)$
→$(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)$
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where $p, q$ are integers and $\text{q}\neq0.$
$0.40\overline{7}$
→$0.40\overline{7}$