goals

For the following system of equation determine the value of k for which the given system of equation has a unique solution.

$$2x-3y=1$$

$$kx+5y=7$$

$$2x-3y=1$$

$$kx+5y=7$$

$$3(5m-7)-2(9m-11)=4(8m-13)-17$$

Let $$Ax+By=C$$ and $$A'x+B'y=C'$$ represent two lines.

(a) If $$\displaystyle \frac{A}{A'} = \frac{B}{B'} \neq \frac{C}{C'}$$, then the lines are parallel.

Above properties provide us a way to write the equation of a line parallel or perpendicular to a given line that contains a given point not on the line. For example, suppose that we want the equation of the line pependicular to $$3x+4y=6$$ that contains the point (1, 2). The form $$4x-3y=k$$, where k is a constant, represents a family of lines perpendicular to $$3x+4y=6$$ because we have satisfied the condition $$AA' = - BB'$$. Therefore, to find the specific line of the family containing (1, 2), we substitute 1 for x and 2 for y to determine k.

$$4x-3y=k$$

$$4(1)- 3(2)=k$$

$$-2 = k$$

Thus, the equation for the desired line is $$4x-3y=-2$$

(a) If $$\displaystyle \frac{A}{A'} = \frac{B}{B'} \neq \frac{C}{C'}$$, then the lines are parallel.

Above properties provide us a way to write the equation of a line parallel or perpendicular to a given line that contains a given point not on the line. For example, suppose that we want the equation of the line pependicular to $$3x+4y=6$$ that contains the point (1, 2). The form $$4x-3y=k$$, where k is a constant, represents a family of lines perpendicular to $$3x+4y=6$$ because we have satisfied the condition $$AA' = - BB'$$. Therefore, to find the specific line of the family containing (1, 2), we substitute 1 for x and 2 for y to determine k.

$$4x-3y=k$$

$$4(1)- 3(2)=k$$

$$-2 = k$$

Thus, the equation for the desired line is $$4x-3y=-2$$

Simplify $$\dfrac{x}{{\sqrt {x + 4} }},\,\,x > 0$$

Solve:

$$2A - B = - 2C$$

$$ {A - B = - C} $$

$$2A - B = - 2C$$

$$ {A - B = - C} $$

Find A and B.