Question

Find whether the given statement is true or false:
(i) x = 2, y = 3 is a solution of the equation 5x − 3y = 1.
(ii) y = 2x + 5 is a straight line passing through the point (1, 5).
(iii) The area bounded by the line x + y = 6, the x-axis and the y-axis is 18 sq units.

Answer 1

(i) Given equation: $5x-3y=1$.
To check wether $x=2,y=3$ is the solution of the equation.
Putting $x=2,y=3$ in the equation, we get:
$LHS=5x-3y=5\times 2-3\times 3=10-9=1=RHS$
Hence, $x=2,y=3$ is the solution of the equation.
Hence, the statement is true.

(ii) Given equation: $y=2x+5$
When $x=-3$, then $y=-1$.
When $x=-2$, then $y=1$.
When $x=-1$, then $y=3$.
When $x=0$, then $y=5$.
When $x=1$, then $y=7$.

Plot these points on a graph paper and join them in both the directions.


From here, we see that for $x=1$ we have $y=7$≠ 5. Moreover, from the graph we see that $(1,5)$ does not lie on the line $y=2x+5$.
Hence, the statement is false.

(iii) Given : $x+y=6$.
When $x=0$, then $y=6$.
When $y=0$, then $x=6$.
Plotting these points on the graph paper, we get the right-angled triangle$AOB$, such that $AO=6$ unit and $OB=6$ unit.


Hence, $areaof△AOB=\frac{1}{2}\times base\times height=\frac{1}{2}\times 6\times 6=18$ sq unit.
Hence, the statement is true.