Solve the following system of linear equations graphically:
4x – 5y – 20=0
3x + 5y – 15 =0
Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis.
Answers
Answer:
hope it helps you mate..
Given:
4x – 5y – 20 = 0
3x + 5y – 15 = 0
Concept :
To draw the graph of a linear equation in two variable a series of steps to be followed which is given below:
Step I:Obtain the linear equation. Let the equation be ax + by + c = 0
Step II: Express one unknown quantity in terms of other here. Express y in terms of x to get y = - (ax + c /b)
Step III: For any two values of x, calculate the corresponding values of y from the expression in step II to obtain two solutions say (x1, y1 and( x2 ,y2)
Step IV: plot the points (x1, y1 and( x2 ,y2) on graph paper on a suitable scale.
Step V:Draw a line passing through points marked in step IV. The line so obtained is the graph of the equation ax + by + c = 0
Solution :
4x – 5y – 20 passes through (5,0) and (0, -4)
3x + 5y – 15 = 0 passes through (5,0) , (0,3)
Two lines intersect at a point A (5,0) .
Hence x is 5 and y is 0.
The vertices of the triangle ABC formed by the lines representing the above equation are A (5,0) , B (-4,0) and C (0,3
Table and the Graph of the given systems of equations are in the attachment below :
HOPE THIS ANSWER WILL HELP YOU……
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