SSC 10th Maharashtra Board: Mathematics Part 1 (Practice Set 1.2)
Question 1
Complete the following table to draw the graph of the equations:
(I) $x + y = 3$
(II) $x - y = 4$
Solution:
For equation $x + y = 3$:
When $x = 3$, $3 + y = 3 \implies y = 0$.
When $y = 5$, $x + 5 = 3 \implies x = -2$.
When $y = 3$, $x + 3 = 3 \implies x = 0$.
$x$ $3$ $-2$ $0$ $y$ $0$ $5$ $3$ $(x, y)$ $(3, 0)$ $(-2, 5)$ $(0, 3)$ For equation $x - y = 4$:
When $y = 0$, $x - 0 = 4 \implies x = 4$.
When $x = -1$, $-1 - y = 4 \implies y = -5$.
When $x = 0$, $0 - y = 4 \implies y = -4$.
$x$ $4$ $-1$ $0$ $y$ $0$ $-5$ $-4$ $(x, y)$ $(4, 0)$ $(-1, -5)$ $(0, -4)$
Question 2 (i)
Solve the following simultaneous equations graphically:
$x + y = 6$ ; $x - y = 4$
Solution:
Equation 1: $x + y = 6 \implies y = 6 - x$
Points for plotting: $(0, 6)$, $(2, 4)$, $(6, 0)$
Equation 2: $x - y = 4 \implies y = x - 4$
Points for plotting: $(0, -4)$, $(4, 0)$, $(5, 1)$
Plotting these points on a graph paper, we get two straight lines. The lines intersect each other at the point $(5, 1)$.
Therefore, the solution of the given simultaneous equations is $x = 5$, $y = 1$.
Question 2 (ii)
Solve the following simultaneous equations graphically:
$x + y = 5$ ; $x - y = 3$
Solution:
Equation 1: $x + y = 5 \implies y = 5 - x$
Points for plotting: $(0, 5)$, $(2, 3)$, $(5, 0)$
Equation 2: $x - y = 3 \implies y = x - 3$
Points for plotting: $(0, -3)$, $(3, 0)$, $(4, 1)$
The lines representing these equations intersect at the point $(4, 1)$.
Therefore, the solution is $x = 4$, $y = 1$.
Question 2 (iii)
Solve the following simultaneous equations graphically:
$x + y = 0$ ; $2x - y = 9$
Solution:
Equation 1: $x + y = 0 \implies y = -x$
Points for plotting: $(0, 0)$, $(2, -2)$, $(3, -3)$
Equation 2: $2x - y = 9 \implies y = 2x - 9$
Points for plotting: $(0, -9)$, $(3, -3)$, $(4, -1)$
The two straight lines intersect at the point $(3, -3)$.
Therefore, the solution is $x = 3$, $y = -3$.
Question 2 (iv)
Solve the following simultaneous equations graphically:
$3x - y = 2$ ; $2x - y = 3$
Solution:
Equation 1: $3x - y = 2 \implies y = 3x - 2$
Points for plotting: $(0, -2)$, $(1, 1)$, $(-1, -5)$
Equation 2: $2x - y = 3 \implies y = 2x - 3$
Points for plotting: $(0, -3)$, $(1, -1)$, $(-1, -5)$
The lines representing the equations intersect at the point $(-1, -5)$.
Therefore, the solution is $x = -1$, $y = -5$.
Question 2 (v)
Solve the following simultaneous equations graphically:
$3x - 4y = -7$ ; $5x - 2y = 0$
Solution:
Equation 1: $3x - 4y = -7 \implies 4y = 3x + 7 \implies y = \frac{3x + 7}{4}$
Points for plotting: $(-1, 1)$, $(3, 4)$, $(1, 2.5)$
Equation 2: $5x - 2y = 0 \implies 2y = 5x \implies y = \frac{5x}{2}$
Points for plotting: $(0, 0)$, $(2, 5)$, $(1, 2.5)$
The two lines intersect each other at the point $(1, 2.5)$.
Therefore, the solution is $x = 1$, $y = 2.5$.
Question 2 (vi)
Solve the following simultaneous equations graphically:
$2x - 3y = 4$ ; $3y - x = 4$
Solution:
Equation 1: $2x - 3y = 4 \implies 3y = 2x - 4 \implies y = \frac{2x - 4}{3}$
Points for plotting: $(2, 0)$, $(-1, -2)$, $(5, 2)$
Equation 2: $3y - x = 4 \implies 3y = x + 4 \implies y = \frac{x + 4}{3}$
Points for plotting: $(-4, 0)$, $(2, 2)$, $(8, 4)$
By plotting these points, we find that the lines intersect at the point $(8, 4)$.
Therefore, the solution is $x = 8$, $y = 4$.
