23 questions · timed · auto-graded
| x | 3 | -3 | -6 | 6 | 0 |
| y | 4 | -4 | -8 | 8 | 0 |
| x | 0 | 3 | -3 | $\frac{9}{2}$ | 2 |
| y | -2 | 0 | -4 | 1 | $\frac{-2}{3}$ |
(0, -5)
Putting the value in the given equation,
We have:
LHS = 5(0) - 4(-5)
= 0 + 20
= 20
RHS = 20
LHS = RHS
Thus, (0, -5) is a solution of the given equation.
$\frac{2\text{x}}{5}+\frac{3\text{y}}{10}=3$
| x | 0 | $\frac{15}{2}$ | 5 | 10 | 3 |
| y | 10 | 0 | $\frac{10}{3}$ | $\frac{-10}{3}$ | 6 |
$2\text{x}-\frac{\text{y}}{5}+6=0$
$2\text{x}-\frac{\text{y}}{5}+6=0$
$\Rightarrow10\text{x}-\text{y}+30=0$
On comparing this equation with ax + by + c = 0, we obtain a = 10, b = -1 and c = 30$\Big(2,\ \frac{-5}{2}\Big)$
$\Big(2,\ \frac{-5}{2}\Big)$
Putting the value in the given equation, We have: LHS $=5(2)-4\Big(\frac{-5}{2}\Big)$ = 10 + 10 = 20 RHS = 20 LHS = RHS Thus, $\Big(2,\ \frac{-5}{2}\Big)$ is a solution of the given equation.$\frac{\text{x}}{5}-\frac{\text{y}}{6}=1$
$\frac{\text{x}}{5}-\frac{\text{y}}{6}=1$
$\Rightarrow\frac{6\text{x}-5\text{y}}{30}=1$
$\Rightarrow6\text{x}-5\text{y}=30$
$\Rightarrow6\text{x}-5\text{y}-30=0$
On comparing this equation with ax + by + c = 0, we obtain a = 6, b = -5 and c = -30$\sqrt{2}\text{x}+\sqrt{3}\text{y}=5$
$\sqrt{2}\text{x}+\sqrt{3}\text{y}=5$
$\sqrt{2}\text{x}+\sqrt{3}\text{y}-5=0$
On comparing this equation with ax + by + c = 0, we obtain $\text{a}=\sqrt{2},\ \text{b}=\sqrt{3}$ and c = -30$\frac{\text{x}}{2}-\frac{\text{y}}{3}=\frac{1}{6}+\text{y}$
$\frac{\text{x}}{2}-\frac{\text{y}}{3}=\frac{1}{6}+\text{y}$
$\Rightarrow\frac{\text{x}}{2}-\frac{\text{y}}{3}-\text{y}=\frac{1}{6}$
$\Rightarrow\frac{3\text{x}-2\text{y}-6\text{y}}{6}=\frac{1}{6}$
$\Rightarrow3\text{x}-8\text{y}=1$
$\Rightarrow3\text{x}-8\text{y}-1=0$
On comparing this equation with ax + by + c = 0, we obtain a = 3, b = -8 and c = -1$\Big(-2,\ \frac{5}{2}\Big)$
$\Big(-2,\ \frac{5}{2}\Big)$
Putting the value in the given equation, We have: LHS $=5(-2)-4\Big(\frac{5}{2}\Big)$ = -10 - 10 = -20 RHS = 20$\text{LHS}\neq\text{RHS}$
Thus, $\Big(-2,\ \frac{5}{2}\Big)$ is not a solution of the given equation.$\text{LHS}\neq\text{RHS}$
Thus, (0, 5) is not a solution of the given equation.$\Rightarrow3\text{x}+5\text{y}-\frac{15}{2}=0$
⇒ 6x + 10y - 15 = 0 On comparing this equation with ax + by + c = 0, we obtain a = 6, b = 10 and c = -15