QuestionFind the value of $k$ for which the system $-12x + 4y = 40$ and $Ax - 2y = -20$ has no solution.

Studdy Solution

STEP 1

Assumptions1. The system of equations is given by\begin{align} -12x +4y &=40 \\ Ax -y &= -20\end{align} . $k$ is a constant and $x$ and $y$ are variables.
3. We are looking for the value of $k$ that makes the system of equations have no solution.

STEP 2

In a system of two linear equations, the equations represent two lines in a plane. The system has no solution if and only if the two lines are parallel. Two lines are parallel if and only if their slopes are equal.

STEP 3

The slope of a line in the form $Ax + By = C$ is given by $-A/B$ . So, we first need to rewrite the given equations in this form.

STEP 4

Rewrite the first equation in the form $Ax + By = C$ .
$-12x +4y =40$

STEP 5

Rewrite the second equation in the form $Ax + By = C$ .
$Ax -2y = -20$

STEP 6

Now we can calculate the slopes of the two lines. The slope of the first line is $-(-12)/4 =3$ .

STEP 7

The slope of the second line is $-A/(-2) = A/2$ .

STEP 8

For the system to have no solution, the slopes of the two lines must be equal. So, we set the two slopes equal to each other and solve for $A$ .
$3 = A/2$

STEP 9

Multiply both sides of the equation by2 to solve for $A$ .
$A =3 \times2$

STEP 10

Calculate the value of $A$ .
$A =6$ So, for the system of equations to have no solution, $k$ (which is equal to $A$ in this case) must be6.

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QuestionFind the value of $k$ for which the system $-12x + 4y = 40$ and $Ax - 2y = -20$ has no solution.

Studdy Solution

STEP 1

Assumptions1. The system of equations is given by\begin{align} -12x +4y &=40 \\ Ax -y &= -20\end{align} . $k$ is a constant and $x$ and $y$ are variables.
3. We are looking for the value of $k$ that makes the system of equations have no solution.

STEP 2

In a system of two linear equations, the equations represent two lines in a plane. The system has no solution if and only if the two lines are parallel. Two lines are parallel if and only if their slopes are equal.

STEP 3

The slope of a line in the form $Ax + By = C$ is given by $-A/B$ . So, we first need to rewrite the given equations in this form.

STEP 4

Rewrite the first equation in the form $Ax + By = C$ .
$-12x +4y =40$

STEP 5

Rewrite the second equation in the form $Ax + By = C$ .
$Ax -2y = -20$

STEP 6

Now we can calculate the slopes of the two lines. The slope of the first line is $-(-12)/4 =3$ .

STEP 7

The slope of the second line is $-A/(-2) = A/2$ .

STEP 8

For the system to have no solution, the slopes of the two lines must be equal. So, we set the two slopes equal to each other and solve for $A$ .
$3 = A/2$

STEP 9

Multiply both sides of the equation by2 to solve for $A$ .
$A =3 \times2$

STEP 10

Calculate the value of $A$ .
$A =6$ So, for the system of equations to have no solution, $k$ (which is equal to $A$ in this case) must be6.

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QuestionFind the value of kkk for which the system −12x+4y=40-12x + 4y = 40−12x+4y=40 and Ax−2y=−20Ax - 2y = -20Ax−2y=−20 has no solution.

Studdy Solution

STEP 1

Assumptions1. The system of equations is given by\begin{align*} -12x +4y &=40 \\ Ax -y &= -20\end{align*} . kkk is a constant and xxx and yyy are variables.3. We are looking for the value of kkk that makes the system of equations have no solution.

STEP 2

In a system of two linear equations, the equations represent two lines in a plane. The system has no solution if and only if the two lines are parallel. Two lines are parallel if and only if their slopes are equal.

STEP 3

The slope of a line in the form Ax+By=CAx + By = CAx+By=C is given by −A/B-A/B−A/B. So, we first need to rewrite the given equations in this form.

STEP 4

Rewrite the first equation in the form Ax+By=CAx + By = CAx+By=C.−12x+4y=40-12x +4y =40−12x+4y=40

STEP 5

Rewrite the second equation in the form Ax+By=CAx + By = CAx+By=C.Ax−2y=−20Ax -2y = -20Ax−2y=−20

STEP 6

Now we can calculate the slopes of the two lines. The slope of the first line is −(−12)/4=3-(-12)/4 =3−(−12)/4=3.

STEP 7

The slope of the second line is −A/(−2)=A/2-A/(-2) = A/2−A/(−2)=A/2.

STEP 8

For the system to have no solution, the slopes of the two lines must be equal. So, we set the two slopes equal to each other and solve for AAA.3=A/23 = A/23=A/2

STEP 9

Multiply both sides of the equation by2 to solve for AAA.A=3×2A =3 \times2A=3×2

STEP 10

Calculate the value of AAA.A=6A =6A=6So, for the system of equations to have no solution, kkk (which is equal to AAA in this case) must be6.

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QuestionFind the value of $k$ for which the system $-12x + 4y = 40$ and $Ax - 2y = -20$ has no solution.

Assumptions1. The system of equations is given by\begin{align} -12x +4y &=40 \\ Ax -y &= -20\end{align} . $k$ is a constant and $x$ and $y$ are variables.
3. We are looking for the value of $k$ that makes the system of equations have no solution.

The slope of a line in the form $Ax + By = C$ is given by $-A/B$ . So, we first need to rewrite the given equations in this form.

Rewrite the first equation in the form $Ax + By = C$ .
$-12x +4y =40$

Rewrite the second equation in the form $Ax + By = C$ .
$Ax -2y = -20$

Now we can calculate the slopes of the two lines. The slope of the first line is $-(-12)/4 =3$ .

The slope of the second line is $-A/(-2) = A/2$ .

For the system to have no solution, the slopes of the two lines must be equal. So, we set the two slopes equal to each other and solve for $A$ .
$3 = A/2$

Multiply both sides of the equation by2 to solve for $A$ .
$A =3 \times2$

Calculate the value of $A$ .
$A =6$ So, for the system of equations to have no solution, $k$ (which is equal to $A$ in this case) must be6.