### Video Transcript

Find the equation of the straight
line which passes through the point of intersection of the two lines negative four
π₯ plus 15π¦ equals negative 15 and negative four π₯ plus three π¦ equals 14 and is
parallel to the straight line π« equals four, zero plus π times five, negative
four.

The two bits of information we have
about the straight line whose equation we want to solve for then is that it passes
through the point of intersection of these two lines and that itβs parallel to this
line. In general, we can write the
equation of a straight line like this. The π¦-coordinate of a line is
equal to the slope π multiplied by the corresponding π₯-coordinate plus the
π¦-intercept called π. For a given line then, if weβre
able to determine its slope and π¦-intercept, then we have defined that line. When we consider the equation of
the straight line we want to solve for, we can start by finding out its slope
π.

Recall that weβre told that this
line is parallel to this given straight line. A line written this way is said to
be in vector form. When thatβs true, this first vector
represents a point on the line and the second vector, the one here by which weβre
multiplying π, gives a vector thatβs parallel to the line. That is, a vector where we go five
units in the positive π₯-direction and four units in the negative π¦-direction is
parallel to this line. And therefore, itβs parallel to the
line whose equation we want to solve for.

In general, the slope of a line
equals the amount the line changes in the π¦-dimension divided by the corresponding
change in the π₯-dimension. For our line written in vector
form, we see that that ratio is negative four over five. And so we can use this value for π
in the equation of our straight line. The last thing we need then to
specify our lineβs equation is the value of the π¦-intercept π. We can solve for π if we know a
point that this line passes through. Knowing those coordinates, we can
substitute in the π₯- and π¦-value of that pair and then solve this resulting
equation for π.

Recall weβre told that our straight
line passes through the point of intersection of these two lines. If we can find that point then,
weβll have π₯- and π¦-values we can substitute in here to solve for π. Looking at the equations of these
two lines that intersect, we know that because they intersect, there is some π₯- and
π¦-value, which is true for both equations. This means that we have two
independent equations as well as two unknowns to solve for.

We can start solving for these
unknowns by subtracting the second equation from the first. When we do this, for the π₯-value,
we have negative four π₯ minus negative four π₯, that gives us zero, while for the
π¦-value, we have 15π¦ minus three π¦. Thatβs positive 12π¦. And then, lastly, this equals
negative 15 minus 14. Thatβs negative 29. From this, we see that π¦ equals
negative 29 over 12. Thatβs the π¦-coordinate of the
point of intersection of these two lines. To solve for the corresponding
π₯-value, we take this π¦-value and substitute it in to either of these two
equations.

Say that we choose the second
equation for this substitution. That gives us negative four π₯ plus
three times negative 29 over 12 equals 14, or on the left-hand side, negative four
π₯ minus 29 over four. If we add positive 29 over four to
both sides and then divide both sides of that equation by negative four, we get this
expression. And we can then note that 14 is the
same as 56 divided by four and then that 56 plus 29 equals 85 and then 85 over four
divided by negative four equals negative 85 over 16.

This then is the π₯-value of the
point of intersection of the two lines given to us. And since weβve already solved for
the corresponding π¦-value, we can now write out the coordinate pair of that
point. Like we mentioned, weβre able to
use these π₯- and π¦-values in our lineβs equation to solve for π. Substituting those values in gives
us this expression, and if we then multiply negative four-fifths by negative
eighty-five sixteenths, we get a result of 340 over 80. And note that if we divide both the
top and the bottom of this fraction by 10, the zeros effectively cancel out.

We can write then that negative 29
over 12 equals 34 over eight plus π. We can then subtract 34 over eight
from both sides. And note that if we multiply top
and bottom of this fraction by two and we multiply numerator and denominator of this
fraction by three, then both of these fractions will have the same denominator. So we can combine them. Negative 58 minus 102 is negative
160. And then we know that the number
eight divides into 160 exactly 20 times, while it divides into 24 exactly three
times. Our simplified value for π then is
negative 20 over three.

Finally then, we can substitute
this value for π into the equation of our line. And we see that π¦ equals negative
four-fifths π₯ minus 20 over three. This is our answer in
slopeβintercept form, while an alternative answer form is to bring both variables to
the same side of our equation. If we add four-fifths π₯ to both
sides, we get this equivalent expression.

The equation of the line that
satisfies the conditions given in the problem statement is π¦ plus four-fifths π₯
equals negative 20 divided by three.