Solve the following pair of linear equations graphically. Also write the observations.
i) x + y = 1 ; 2 x – 3y = 7
ii) x +2y = 4 ; 2x + 4y = 12
iii) 3x –y =2 ; 9x -3y = 6
Answers
Question 1 :-
(i) x + y = 1 ; 2x - 3y = 7
First equation :- x + y = 1
Case 1 :- Let us consider that x equals 0.
→ x + y = 1
→ 0 + y = 1
→ y = 1
∴ First pair → (0, 1)
Case 2 :- Let us consider that x equals 1.
→ x + y = 1
→ 1 + y = 1
→ y = 1 - 1
→ y = 0
∴ Second pair → (1, 0)
Case 3 :- Let us consider that x equals -1.
→ x + y = 1
→ -1 + y = 1
→ y = 1 + 1
→ y = 2
∴ Third pair → (-1, 2)
Plot the points (0, 1), (1, 0), (-1, 2) on a graph, and join the points, the resulting line represents x + y = 1. [Blue line]
Second equation :- 2x - 3y = 7
→ 2x - 3y = 7
→ 2x - 7 = 3y
→ (2x -7)/3 = y [We'll be substituing values in this equation]
Case 1 :- Let us consider that x equals 5.
→ (2x - 7)/3 = y
→ (2(5) - 7)/3 = y
→ (10 - 7)/3 = y
→ 3/3 = y
→ y = 1
∴ First pair → (5, 1)
Case 2 :- Let us consider that x equals -1.
→ (2x - 7)/3 = y
→ (2(-1) -7)/3 = y
→ (-2 -7)/3 = y
→ -9/3 = y
→ y = -3
∴ Second pair → (-1, -3)
Case 3 :- Let us consider that x equals -4.
→ (2x - 7)/3 = y
→ (2(-4) - 7)/3 = y
→ (-8 -7)/3 = y
→ -15/3 = y
→ y = -5
∴ Third pair → (-4,-5)
Plot the points (5, 1), (-1, -3), (-4,-5) on a graph, and join the points, the resulting line represents 2x - 3y = 7. [Red line]
Observations :-
- From the graph, it's apparent that the two lines intersect at the point (2, -1), therefore the solution is (2, -1).
- The pair of lines are intersecting lines, they contain only one solution, and only have one point in common.
Question 2 :-
(ii) x + 2y = 4 ; 2x + 4y = 12
First equation :- x + 2y = 4
Case 1 :- Let us consider that x equals 2.
→ x + 2y = 4
→ 2 + 2y = 4
→ 2y = 4 - 2
→ 2y = 2
→ y = 1
∴ First pair → (2, 1)
Case 2 :- Let us consider that x equals 0.
→ x + 2y = 4
→ 0 + 2y = 4
→ 2y = 4
→ y = 4/2
→ y = 2
∴ Second pair → (0, 2)
Case 3 :- Let us consider that x equals -2.
→ x + 2y = 4
→ -2 + 2y = 4
→ 2y = 4+2
→ y = 6/2
→ y = 3
∴ Third pair → (-2, 3)
Plot the points (2, 1), (0, 2), (-2, 3) on a graph, and join the points, the resulting line represents x + 2y = 4. [Orange line]
Second equation :- 2x + 4y = 12
→ 2x + 4y = 12
→ 2(x + 2y) = 12
→ (x + 2y) = 12/2
→ x + 2y = 6 [We'll be substituing values in this equation]
Case 1 :- Let us consider that x equals O.
→ x + 2y = 6
→ 0 + 2y = 6
→ 2y = 6
→ y = 6/2
→ y = 3
∴ First pair → (0, 3)
Case 2 :- Let us consider that x equals 2.
→ x + 2y = 6
→ 2 + 2y = 6
→ 2y = 6 - 2
→ y = 4/2
→ y = 2
∴ Second pair → (2, 2)
Case 3 :- Let us consider that y equals 4.
→ x + 2y = 6
→ x + 2(4) = 6
→ x + 8 = 6
→ x = 6 - 8
→ x = -2
∴ Third pair → (-2, 4)
Plot the points (0, 3), (2, 2), (-2, 4) on a graph, and join the points, the resulting line represents 2x + 4y = 12. [Black line]
Observations :-
- From the graph, it's apparent that the two lines are parallel to each other, and do not intersect. Therefore, there is no solution.
- The pair of lines are parallel lines, therefore they do not contain any solutions, and do not hold any points in common.
Question 3 :-
(iii) 3x - y = 2 ; 9x - 3y = 6
First equation :- 3x - y = 2
Case 1 :- Let us consider that x equals O.
→ 3x - y = 2
→ 3(0) - y = 2
→ - y = 2
→ y = -2
∴ First pair → (0, -2)
Case 2 :- Let us consider that x equals 1.
→ 3x - y = 2
→ 3(1) - y = 2
→ 3 - y = 2
→ -y = 2 - 3
→ -y = -1
→ y = 1
∴ Second pair → (1, 1)
Case 3 :- Let us consider that x equals 2.
→ 3x - y = 2
→ 3(2) - y = 2
→ 6 - y = 2
→ -y = 2 - 6
→ -y = -4
→ y = 4
∴ Third pair → (2, 4)
Plot the points (0, -2), (1, 1), (2, 4) on a graph, and join the points, the resulting line represents x + y = 1. [Blue line]
Second equation :- 9x - 3y = 6
→ 9x - 3y = 6
→ 3(3x - y) = 6
→ 3x - y = 6/3
→ 3x - y = 2
→ 3x - y = 2
This equation is the same as the first equation we've found coordinates for, so we can make use of the same three points for plotting purposes, which are (0, -2), (1, 1) and (2, 4). [Blue line]
Observations :-
- From the graph, it's apparent that the two lines are co-incident lines, they have infinitely many solutions.
- The pair of lines are co-incident, and they contain only infinitely many solutions and have infinite points in common.
[Graphs attached, made with Desmos]
