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Post a LessonAnswered on 16 Apr Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Pair of linear equations in two variable
Sadika
To have an infinite number of solutions for the given pair of equations, the equations must represent the same line. This means that the second equation is a multiple of the first equation, or vice versa.
Let's analyze the equations:
Answered on 16 Apr Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Pair of linear equations in two variable
Sadika
To solve the pair of equations graphically, let's first rewrite them in slope-intercept form:&space;\(3y&space;=&space;-x&space;+&space;6\)&space;\(y&space;=&space;-\frac{1}{3}x&space;+&space;2\)&space;2.&space;\(2x&space;-&space;3y&space;=&space;12\)&space;\(3y&space;=&space;2x&space;-&space;12\)&space;\(y&space;=&space;\frac{2}{3}x&space;-&space;4\)&space;Now,&space;let's&space;plot&space;these&space;lines&space;on&space;a&space;graph:&space;&space;To&space;find&space;the&space;area&space;of&space;the&space;triangle&space;formed&space;by&space;these&space;lines&space;and&space;the&space;y-axis,&space;we&space;need&space;to&space;find&space;the&space;x-intercepts&space;of&space;both&space;lines.&space;For&space;the&space;first&space;line:&space;\[y&space;=&space;-\frac{1}{3}x&space;+&space;2\]&space;\[0&space;=&space;-\frac{1}{3}x&space;+&space;2\]&space;\[\frac{1}{3}x&space;=&space;2\]&space;\[x&space;=&space;6\]&space;For&space;the&space;second&space;line:&space;\[y&space;=&space;\frac{2}{3}x&space;-&space;4\]&space;\[0&space;=&space;\frac{2}{3}x&space;-&space;4\]&space;\[\frac{2}{3}x&space;=&space;4\]&space;\[x&space;=&space;6\]&space;Now,&space;we&space;have&space;the&space;x-coordinate&space;of&space;the&space;intersection&space;point:&space;\(x&space;=&space;6\). "1. \(x + 3y=6\) \(3y = -x + 6\) \(y = -\frac{1}{3}x + 2\) 2. \(2x - 3y=12\) \(3y = 2x - 12\) \(y = \frac{2}{3}x - 4\) Now, let's plot these lines on a graph:  To find the area of the triangle formed by these lines and the y-axis, we need to find the x-intercepts of both lines. For the first line: \[y = -\frac{1}{3}x + 2\] \[0 = -\frac{1}{3}x + 2\] \[\frac{1}{3}x = 2\] \[x = 6\] For the second line: \[y = \frac{2}{3}x - 4\] \[0 = \frac{2}{3}x - 4\] \[\frac{2}{3}x = 4\] \[x = 6\] Now, we have the x-coordinate of the intersection point: \(x = 6\).")
Answered on 16 Apr Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression
Sadika
To show that the above equation
Answered on 16 Apr Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression
Sadika
To find the common difference (d) of an arithmetic progression (AP), you can use the formula:
Answered on 16 Apr Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression
Sadika
Let's denote the first term of the arithmetic progression (AP) as a and the common difference as d
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