The system of equations 2x + y - 5 = 0,x - 2y | vedclass

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(D) For the system of equations to be consistent,all three lines must intersect at a single point or be concurrent. First,solve the first two equation

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$-16$

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(D) For the system of equations to be consistent,all three lines must intersect at a single point or be concurrent. First,solve the first two equations: $2x + y = 5$  $(1)$ $x - 2y = -1$  $(2)$ Multiply equation $(1)$ by $2$: $4x + 2y = 10$  $(3)$ Adding $(2)$ and $(3)$: $5x = 9 \implies x = \frac{9}{5}$ Substitute $x$ into $(1)$: $2(\frac{9}{5}) + y = 5 \implies y = 5 - \frac{18}{5} = \frac{7}{5}$ Since the system is consistent,the point $(\frac{9}{5}, \frac{7}{5})$ must satisfy the third equation $2x - 14y - a = 0$: $2(\frac{9}{5}) - 14(\frac{7}{5}) - a = 0$ $\frac{18}{5} - \frac{98}{5} = a$ $a = -\frac{80}{5} = -16$

The system of equations $2x + y - 5 = 0$,$x - 2y + 1 = 0$,and $2x - 14y - a = 0$ is consistent. Then,$a$ is equal to

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