Practice Set 1.2 Linear Equations In Two Variables Class 10th Mathematics Part 1 MHB Solution (49)

Practice Set 1.2 Linear Equations In Two Variables

Practice Set 1.2 Linear Equations In Two Variables
Class 10th Mathematics Part 1 MHB Solution (49)

Practice Set 1.2

Question 1.

Complete the following table to draw graph of the equations -

(I) \(x+y = 3\)

(II) \(x - y = 4\)

Answer:

(1). For Equation (I): \(x+y = 3\)

i. If \(x=3\):
\( \quad 3 + y = 3 \)
\( \quad y = 3 - 3 \)
\( \quad y = 0 \)
Point: \((3, 0)\)
ii. If \(x=-2\):
\( \quad -2 + y = 3 \)
\( \quad y = 3 + 2 \)
\( \quad y = 5 \)
Point: \((-2, 5)\)
iii. If \(x=0\):
\( \quad 0 + y = 3 \)
\( \quad y = 3 \)
Point: \((0, 3)\)

Table for \(x+y=3\):

xy(x,y)
30(3,0)
-25(-2,5)
03(0,3)

Graphical Representation for \(x+y=3\):

Graph for x+y=3

(2). For Equation (II): \(x-y = 4\)

i. If \(x=4\):
\( \quad 4 - y = 4 \)
\( \quad -y = 4 - 4 \)
\( \quad -y = 0 \)
\( \quad y = 0 \)
Point: \((4, 0)\)
ii. If \(x=-1\):
\( \quad -1 - y = 4 \)
\( \quad -y = 4 + 1 \)
\( \quad -y = 5 \)
\( \quad y = -5 \)
Point: \((-1, -5)\)
iii. If \(x=0\):
\( \quad 0 - y = 4 \)
\( \quad -y = 4 \)
\( \quad y = -4 \)
Point: \((0, -4)\)

Table for \(x-y=4\):

xy(x,y)
40(4,0)
-1-5(-1,-5)
0-4(0,-4)

Graphical Representation for \(x-y=4\):

Graph for x-y=4

Question 2.

Solve the following simultaneous equation graphically.

(1) \(x + y = 6\); \(x - y = 4\)

Answer:

Equation I: \(x + y = 6\)

If \(x=0\), \(0+y=6 \Rightarrow y=6\). Point: \((0,6)\)
If \(x=6\), \(6+y=6 \Rightarrow y=0\). Point: \((6,0)\)
If \(x=5\), \(5+y=6 \Rightarrow y=1\). Point: \((5,1)\)
If \(x=3\), \(3+y=6 \Rightarrow y=3\). Point: \((3,3)\)
xy(x,y)
06(0,6)
60(6,0)
51(5,1)
33(3,3)

Equation II: \(x - y = 4\)

If \(x=0\), \(0-y=4 \Rightarrow y=-4\). Point: \((0,-4)\)
If \(x=4\), \(4-y=4 \Rightarrow y=0\). Point: \((4,0)\)
If \(x=2\), \(2-y=4 \Rightarrow -y=2 \Rightarrow y=-2\). Point: \((2,-2)\)
If \(x=5\), \(5-y=4 \Rightarrow -y=-1 \Rightarrow y=1\). Point: \((5,1)\)
xy(x,y)
0-4(0,-4)
40(4,0)
2-2(2,-2)
51(5,1)

Graphical Representation:

Graphical solution for Question 2

Calculating intersecting point (Algebraically):

Given equations are:

\(x + y = 6\) --- (I)

\(x - y = 4\) --- (II)

Adding Equation (I) and Equation (II):

\((x+y) + (x-y) = 6 + 4\)

\(2x = 10\)

\(x = \frac{10}{2}\)

\(x = 5\)

Substituting \(x=5\) into Equation (I):

\(5 + y = 6\)

\(y = 6 - 5\)

\(y = 1\)

Intersection Point: \((5,1)\)

Question 3.

Solve the following simultaneous equation graphically.

\(x + y = 5\); \(x - y = 3\)

Answer:

Equation I: \(x + y = 5\)

If \(x=0\), \(y=5\). Point: \((0,5)\)
If \(x=5\), \(y=0\). Point: \((5,0)\)
If \(x=2\), \(y=3\). Point: \((2,3)\)
If \(x=4\), \(y=1\). Point: \((4,1)\)
xy(x,y)
05(0,5)
50(5,0)
23(2,3)
41(4,1)

Equation II: \(x - y = 3\)

If \(x=0\), \(y=-3\). Point: \((0,-3)\)
If \(x=3\), \(y=0\). Point: \((3,0)\)
If \(x=2\), \(y=-1\). Point: \((2,-1)\)
If \(x=4\), \(y=1\). Point: \((4,1)\)
xy(x,y)
0-3(0,-3)
30(3,0)
2-1(2,-1)
41(4,1)

Graphical Representation:

Graphical solution for Question 3

Calculating intersecting point (Algebraically):

\(x + y = 5\) --- (I)

\(x - y = 3\) --- (II)

Adding Equation (I) and Equation (II):

\((x+y) + (x-y) = 5 + 3\)

\(2x = 8\)

\(x = 4\)

Substituting \(x=4\) into Equation (I):

\(4 + y = 5\)

\(y = 1\)

Intersection Point: \((4,1)\)

Question 4.

Solve the following simultaneous equation graphically.

\(x + y = 0\); \(2x - y = 9\)

Answer:

Equation I: \(x + y = 0 \Rightarrow y = -x\)

If \(x=0\), \(y=0\). Point: \((0,0)\)
If \(x=1\), \(y=-1\). Point: \((1,-1)\)
If \(x=-2\), \(y=2\). Point: \((-2,2)\)
If \(x=3\), \(y=-3\). Point: \((3,-3)\)
xy(x,y)
00(0,0)
1-1(1,-1)
-22(-2,2)
3-3(3,-3)

Equation II: \(2x - y = 9 \Rightarrow y = 2x - 9\)

If \(x=0\), \(y=-9\). Point: \((0,-9)\)
If \(x=4.5\), \(y=0\). Point: \((4.5,0)\)
If \(x=4\), \(y=-1\). Point: \((4,-1)\)
If \(x=3\), \(y=-3\). Point: \((3,-3)\)
xy(x,y)
0-9(0,-9)
4.50(4.5,0)
4-1(4,-1)
3-3(3,-3)

Graphical Representation:

Graphical solution for Question 4

Calculating intersecting point (Algebraically):

From Equation I, \(y = -x\).

Substitute \(y = -x\) into Equation II: \(2x - (-x) = 9\)

\(2x + x = 9\)

\(3x = 9\)

\(x = 3\)

Substitute \(x=3\) into Equation I: \(3 + y = 0\)

\(y = -3\)

Intersection Point: \((3,-3)\)

Question 5.

Solve the following simultaneous equation graphically.

\(3x - y = 2\); \(2x - y = 3\)

Answer:

Equation I: \(3x - y = 2 \Rightarrow y = 3x - 2\)

If \(x=0\), \(y=-2\). Point: \((0,-2)\)
If \(x=1\), \(y=1\). Point: \((1,1)\)
If \(x=-1\), \(y=-5\). Point: \((-1,-5)\)
If \(x=2/3\), \(y=0\). Point: \( (2/3, 0) \approx (0.67,0)\)
xy(x,y)
0-2(0,-2)
11(1,1)
-1-5(-1,-5)
2/30(0.67,0)

Equation II: \(2x - y = 3 \Rightarrow y = 2x - 3\)

If \(x=0\), \(y=-3\). Point: \((0,-3)\)
If \(x=1.5\), \(y=0\). Point: \((1.5,0)\)
If \(x=1\), \(y=-1\). Point: \((1,-1)\)
If \(x=-1\), \(y=-5\). Point: \((-1,-5)\)
xy(x,y)
0-3(0,-3)
1.50(1.5,0)
1-1(1,-1)
-1-5(-1,-5)

Graphical Representation:

Graphical solution for Question 5

Calculating intersecting point (Algebraically):

\(3x - y = 2\) --- (I)

\(2x - y = 3\) --- (II)

Subtracting Equation (II) from Equation (I):

\((3x - y) - (2x - y) = 2 - 3\)

\(3x - y - 2x + y = -1\)

\(x = -1\)

Substituting \(x=-1\) into Equation (I):

\(3(-1) - y = 2\)

\(-3 - y = 2\)

\(-y = 5\)

\(y = -5\)

Intersection Point: \((-1,-5)\)

Question 6.

Solve the following simultaneous equation graphically.

\(3x - 4y = -7\); \(5x - 2y = 0\)

Answer:

Equation I: \(3x - 4y = -7 \Rightarrow y = \frac{3x+7}{4}\)

If \(x=1\), \(y = \frac{3(1)+7}{4} = \frac{10}{4} = 2.5\). Point: \((1, 2.5)\)
If \(x=-1\), \(y = \frac{3(-1)+7}{4} = \frac{4}{4} = 1\). Point: \((-1, 1)\)
If \(x=0\), \(y = \frac{3(0)+7}{4} = \frac{7}{4} = 1.75\). Point: \((0, 1.75)\)
If \(y=0\), \(3x = -7 \Rightarrow x = -7/3 \approx -2.33\). Point: \( (-7/3, 0) \))
xy(x,y)
12.5(1, 2.5)
-11(-1, 1)
01.75(0, 1.75)
-7/30(-2.33, 0)

Equation II: \(5x - 2y = 0 \Rightarrow y = \frac{5}{2}x\)

If \(x=0\), \(y = 0\). Point: \((0,0)\)
If \(x=1\), \(y = \frac{5}{2}(1) = 2.5\). Point: \((1, 2.5)\)
If \(x=2\), \(y = \frac{5}{2}(2) = 5\). Point: \((2,5)\)
If \(x=-2\), \(y = \frac{5}{2}(-2) = -5\). Point: \((-2, -5)\)
xy(x,y)
00(0,0)
12.5(1, 2.5)
25(2,5)
-2-5(-2,-5)

Graphical Representation (Plotting both the graphs we get):

Graphical solution for Question 6

Calculating intersecting point (Algebraically):

From Equation II, \(y = \frac{5}{2}x\).

Substitute into Equation I: \(3x - 4\left(\frac{5}{2}x\right) = -7\)

\(3x - 2(5x) = -7\)

\(3x - 10x = -7\)

\(-7x = -7\)

\(x = 1\)

Substitute \(x=1\) into \(y = \frac{5}{2}x\):

\(y = \frac{5}{2}(1)\)

\(y = 2.5\)

Intersection Point: \((1, 2.5)\)