Question 93

The graph of the equation x — 7y = —42, intersects the y-axis at $$P\left(\alpha,\beta\right)$$ and the graph of 6x + y - 15 = 0, intersects the x-axis at $$Q\left(\gamma,\delta\right)$$, What is the value of $$\alpha+\beta+\gamma+\delta?$$

Solution

The graph of the equation x — 7y = —42, intersects the y-axis at $$P\left(\alpha,\beta\right)$$
So, x = 0
0 - 7y = -42
y = 6
$$\alpha$$ = 0
$$\beta$$ = 6
graph of 6x + y - 15 = 0, intersects the x-axis at $$Q\left(\gamma,\delta\right)$$
So, y = 0
6x - 15 = 0
x = 5/2
$$\gamma$$ = 5/2
$$\delta$$ = 0
Now,
$$\alpha+\beta+\gamma+\delta$$
= 0 + 6 + 5/2 + 0 = $$\frac{17}{2}$$


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