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Using a scale of 1cm to 1unit for both t...

Using a scale of 1cm to 1unit for both the axes Draw the graph of the following equations : `6y=5x+10, y=5x-15` From the graph find:
The co-ordinates of the points where the two lines intersect.

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To solve the problem of graphing the equations \(6y = 5x + 10\) and \(y = 5x - 15\) and finding their intersection point, we can follow these steps: ### Step 1: Rearranging the Equations First, we need to rearrange the first equation into the slope-intercept form \(y = mx + b\). 1. Start with the equation: \[ 6y = 5x + 10 \] 2. Divide every term by 6: \[ y = \frac{5}{6}x + \frac{10}{6} \] 3. Simplify \(\frac{10}{6}\) to \(\frac{5}{3}\): \[ y = \frac{5}{6}x + \frac{5}{3} \] The second equation is already in slope-intercept form: \[ y = 5x - 15 \] ### Step 2: Finding Points for the First Equation We will find two points for the first equation \(y = \frac{5}{6}x + \frac{5}{3}\). 1. Let \(x = 0\): \[ y = \frac{5}{6}(0) + \frac{5}{3} = \frac{5}{3} \] Point: \((0, \frac{5}{3})\) 2. Let \(y = 0\): \[ 0 = \frac{5}{6}x + \frac{5}{3} \] Rearranging gives: \[ \frac{5}{6}x = -\frac{5}{3} \] Multiplying both sides by \(\frac{6}{5}\): \[ x = -2 \] Point: \((-2, 0)\) ### Step 3: Finding Points for the Second Equation Now we will find two points for the second equation \(y = 5x - 15\). 1. Let \(x = 4\): \[ y = 5(4) - 15 = 20 - 15 = 5 \] Point: \((4, 5)\) 2. Let \(y = 0\): \[ 0 = 5x - 15 \] Rearranging gives: \[ 5x = 15 \implies x = 3 \] Point: \((3, 0)\) ### Step 4: Plotting the Points Now we plot the points on a graph using a scale of 1 cm = 1 unit. - For the first equation \(y = \frac{5}{6}x + \frac{5}{3}\): - Plot the points \((0, \frac{5}{3})\) and \((-2, 0)\). - For the second equation \(y = 5x - 15\): - Plot the points \((4, 5)\) and \((3, 0)\). ### Step 5: Drawing the Lines Draw straight lines through the points plotted for each equation. ### Step 6: Finding the Intersection Look for the point where the two lines intersect. From the graph, estimate the coordinates of the intersection point. ### Conclusion From the graph, the intersection point is approximately \((4.2, 5.5)\).
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