Question
If $ \displaystyle \begin{vmatrix} \text{x} &\text{amp;}\text{ y} \\ 1 &\text{amp; } 6 \end{vmatrix}= \displaystyle \begin{vmatrix} 1&\text{amp; }8 \\ 1 &\text{amp; } 6 \end{vmatrix}$ then $\text{x}+2\text{y}=$
- 9
- 17
- 10
- 7
Solution:
x = 1.y = 8
$\therefore$ x + 2y = 17
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($A$) intersects $y=x+2$ exactly at one point
($B$) intersects $y=x+2$ exactly at two points
($C$) intersects $y=(x+2)^2$
($D$) does $NOT$ intersect $y=(x+3)^2$