PDF chapter test TRY NOW
Let us consider the system of linear equations and find the solution graphically.
Find the solution to the system of equations \(2x+y=8\) and \(4x+2y=10\) graphically.
Solution:
Let us consider the equation \(2x+y=8\). Now, we shall substitute the values for \(x\) to find the value of \(y\).
When \(x=1\), \(2(1)+y=8\)\(\Rightarrow y=8-2\)\(\Rightarrow y=6\)
When \(x=2\), \(2(2)+y=8\)\(\Rightarrow y=8-4\)\(\Rightarrow y=4\)
When \(x=3\), \(2(3)+y=8\)\(\Rightarrow y=8-6\)\(\Rightarrow y=2\)
When \(x=4\), \(2(4)+y=8\)\(\Rightarrow y=8-8\)\(\Rightarrow y=0\)
Now, writing these values in the table, we have:
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
| \(y\) | \(6\) | \(4\) | \(2\) | \(0\) |
Similarly, plotting the points for the equation \(4x+2y=10\), we have:
When \(x=0\), \(4(0)+2y=10\)\(\Rightarrow 2y=10\)\(\Rightarrow y=5\)
When \(x=1\), \(4(1)+2y=10\)\(\Rightarrow 2y=10-4=6\)\(\Rightarrow y=3\)
When \(x=2\), \(4(2)+2y=10\)\(\Rightarrow 2y=10-8=2\)\(\Rightarrow y=1\)
When \(x=3\), \(4(3)+2y=10\)\(\Rightarrow 2y=10-12=-2\)\(\Rightarrow y=-1\)
Now, writing these values in the table, we have:
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
| \(y\) | \(5\) | \(3\) | \(1\) | \(-1\) |
Now, plotting these points in the graph, we have:
Since the two lines in the graph do not intersect at any point, the graph is said to be an inconsistent system and has no solution.
Another method:
Consider writing the two equations \(2x+y=8\) and \(4x+2y=10\) in the form of \(y = mx + c\) and determine the slope.
Solution:
The slope of the equation \(2x + y = 8\) is given by:
\(y = -2x + 8\)
The slope of the equation \(2x + y = 8\) is \(-2\).
The slope of the equation \(4x + 2y = 10\) is given by:
\(2y = -4x + 10\)
\(y = -2x + 5\)
The slope of the equation \(4x + 2y = 10\) is \(-2\).
Important!
When writing these two equations in the form of \(y = mx+c\), both the equations have the same slope. This proves that the two lines are parallel and does not have a point of intersection.
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