### Video Transcript

Find the equation of the straight
line which passes through the point of intersection of the two lines negative 13π₯
minus five π¦ equals 14 and two π₯ plus 15π¦ equals negative 11 and is parallel to
the straight line π₯ plus eight π¦ equals negative 14.

Okay, so here we have the equation
of these two lines. And weβre told that they intersect
at some π₯π¦-coordinate point. We want to solve for the equation
of the line that passes through that intersection point and is also parallel to this
straight line.

In general, one way to write the
equation of a straight line is to say that its π¦-coordinate is equal to the slope
of the line π multiplied by its π₯-coordinate plus its π¦-intercept π. Since the values of π¦ and π₯ can
range all over the coordinate plane, the equation of a given line is specified by
defining its slope π and its π¦-intercept π.

In the linear equation we want to
find, we can start out by solving for this lineβs slope. Weβll do this by considering the
fact that the line whose equation weβre solving for is parallel to this line. Note that if we take this equation
and then subtract π₯ from both sides, that gives us this result. And if we then divide both sides by
eight, canceling that factor on the left, we find that π¦ equals negative 14 minus
π₯ all divided by eight or, written another way, negative π₯ over eight minus
seven-fourths.

Weβre interested in this form of
our expression because notice that it matches our ππ₯ plus π format. The slope of this line then is the
value by which π₯ is being multiplied. Written out completely, this is
negative one-eighth. So then, this is the slope of this
line. And since the line is parallel to
the line whose equation we want to solve for, we know that these two lines have the
same slope. The updated form of our lineβs
equation then is that π¦ equals negative one-eighth π₯ plus π.

To help us solve for π, weβll want
to know just where it is that these two given lines intersect. The coordinates of that point will
give us π₯- and π¦-values to substitute into this expression to solve for π. We can look at these two equations
as a system of equations. We have two unknowns, π₯ and π¦,
and two independent equations. And if we solve for the π₯- and
π¦-values that satisfy both of these equations at the same time, then weβll have the
coordinates of their point of intersection.

Looking at the values that multiply
π₯ and π¦ in both equations, we see that if we multiply this top equation by
positive three, then that equation becomes negative 39π₯ minus 15π¦ equals 42. This is useful to us because if we
then add these two equations together, we get negative 37π₯ plus zero π¦ equals
31. This implies that π₯ equals
negative 31 over 37. And this then is the π₯-value of
the point of intersection of our two given lines.

To find the corresponding π¦-value
at that point, we can take this π₯-value and substitute it in for π₯ in either one
of these equations. Making that substitution into our
second equation gives us this result, which we can now solve for π¦. Two times negative 31 over 37
equals negative 62 over 37. And if we then add 62 over 37 to
both sides of the equation and divide both sides by 15, we find that π¦ equals
negative 11 plus sixty-two thirty-sevenths all divided by 15.

Another way to write negative 11 is
as negative 407 divided by 37. We get 407 when we multiply 11 by
37. And this means that when we add
this result to 62 over 37, we get negative 345 divided by 37. 15 goes into 345 exactly 23
times. So this fraction overall simplifies
to negative 23 over 37. The point at which our two lines
intersect then has an π₯-coordinate of negative 31 over 37 and a π¦-coordinate of
negative 23 over 37. And because this point lies along
the straight line whose equation we want to solve for, we can substitute these π₯-
and π¦-values into that lineβs equation to solve for the π¦-intercept π. That gives us this expression. And if we multiply these two
fractions together, we get positive 31 over 296. And subtracting this fraction from
both sides gives us this result.

Weβd like to add these two
fractions together. And we can do so by finding a
common denominator. Weβve already seen that 37 when
multiplied by eight gives 296. So we can then multiply both
numerator and denominator of this first fraction by eight. We have then negative 184 over 296
minus 31 over 296, which comes out to negative 215 over 296. This is our value for π, the
π¦-intercept of the equation of the straight line we want to solve for.

So, returning to the equation of
that line, if we substitute in this value for π, then we have that π¦ equals
negative π₯ over eight minus 215 over 296. This is our answer in slope
intercept form. And we can also write this result
with the variables π₯ and π¦ on the same side. If we add π₯ plus eight to both
sides of this equation, then we have the answer weβll give as the equation of this
straight line. π¦ plus π₯ over eight equals
negative 215 over 296. Note that this is the same as π¦
plus one over eight times π₯ equals negative 215 over 296.