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This is one of our many quizzes on Graphing systems of equations. Click on this to access more such competitive quizzes on Algebra.

Graphing systems of equations Quiz 2

mathematics

 Which of the following systems of equations are not consistent??

 Consistent system of equations means having at least one solution.

Consistent system of equations has at least one solution. $ \\ $ For options (a), (b) and (d), we have, $ \\ \dfrac{a}{d} = \dfrac{b}{e}$ but not equal to $\dfrac{c}{f} \\$  Therefore, these system of equations have zero solutions.

 If graphs of two linear equations have same slope and at least one solution, then

 Use the same slope condition to check for the conditions of having at least one solution.

Let $ax + by + c = 0$ and $dx + ey + f = 0$ be the two equations. $ \\ $ Then, slope is same, i.e. $ \\ - \dfrac{a}{b} =  - \dfrac{d}{e} \Rightarrow \dfrac{a}{d} = \dfrac{b}{e} \\ $ The lines are either parallel or overlapping. $ \\ $ Since there exists at least one solution, the lines cannot be parallel.

 The graph of the equation $x + y = 3$ will intersect the graph of ${x^2} + {y^2} = 2$ at how many points?

 Use substitution method to find the intersecting points.

$x + y = 3 \Rightarrow y = 3 - x \\ $ Substitute this in  ${x^2} + {y^2} = 2$, $ \\ {x^2} + {\left( {3 - x} \right)^2} = 2 \\   \Rightarrow {x^2} + {x^2} + 9 - 6x = 2 \\   \Rightarrow 2{x^2} - 6x + 7 = 0 \\  $ Since ${b^2} - 4ac = {6^2} - 4 \times 2 \times 7 = 36 - 56 =  - 20 < 0$, this equation does not have any real solution. $ \\ $ Thus, the graphs of given two equations do not intersect.

 Which of the following graphs represent the graphing of $2x + 3y = 7$ and $2x + y = 3$ ?

 Use the given solutions in options to check which of them satisfies both the given equations.

The solution $\left( {0.5,2} \right)$ satisfies both the given equations. $ \\ $ Also the y-intercept of the equations, i.e. $\dfrac{7}{3}$ and $3$ can be seen to be true in the graph in option (a).

 The orbit of a planet is given by the equation ${x^2} + {y^2} = 5$. An asteroid is moving in the curve given by ${y^2} = kx + 9$. If the path of the asteroid does not cross the orbit of the planet, the possible values of $k$ is

 Use substitution method to get intersecting points and then use the condition for no real solution to exist.

First we will try to find the intersecting points in terms of $k$. We can use a substitution method for this. $ \\  {x^2} + {y^2} = 5 \\   \Rightarrow {x^2} + kx + 9 = 5 \\   \Rightarrow {x^2} + kx + 4 = 0 \\   \Rightarrow x = \dfrac{{ - k \pm \sqrt {{k^2} - 4 \times 4} }}{2} = \dfrac{{ - k \pm \sqrt {{k^2} - 16} }}{2} \\  $ Since there should be no real intersecting point, $ \\  {k^2} - 16 < 0 \\   \Rightarrow \left( {k - 4} \right)\left( {k + 4} \right) < 0 \\   \Rightarrow  - 4 < k < 4 \\  $