Question 670856
basically you're talking about the intersection of 2 lines whose equations are:
4x + 3y - 13 = 0
2x - y - 4 = 0
the fact that these lines are altitudes to sides of a triangle doesn't change the fact that you are simply looking for the point of their intersection.
if there was additional information they were looking for, it would make a difference, but the way the problem is structured, all that other information is extraneous to the heart of the problem which is to find the intersection of the 2 lines.
the 2 equations are, once again:
4x + 3y - 13 = 0
2x - y - 4 = 0
you need to solve these equations simultaneously to get the solution.
you need to first transform them to standard form which is ax + by = c
4x + 3y - 13 = 0 becomes 4x + 3y = 13 once you convert it.
2x - y - 4 = 0 becomes 2x - y = 4 once you convert it.
your 2 equations are now:
4x + 3y = 13
2x - y = 4
we will solve by elimination.
multiply the second equation by 2 to get:
4x - 2y = 8
your 2 equations are now:
4x + 3y = 13
4x - 2y = 8
subtract the second equation from the first equation to get:
5y = 5
divide both sides of this equation by 5 to get:
y = 1
substitute for y in either original equation to find the value of x.
we'll use:
4x + 3y = 13
replace y with 1 to get:
4x + 3 = 13
subtract 3 from both sides of this equation to get:
4x = 10
divide both sides of this equation by 4 to get:
x = 2.5
your solution should be:
x = 2.5
y = 1
this solution should apply to both equation.
the first original equation is:
4x + 3y = 13
replace x with 2.5 and y with 1 to gtet:
4(2.5) + 3(1) = 13 which becomes:
10 + 3 = 13 which is true.
the second original equation is:
2x - y = 4
replace x with 2.5 and y with 1 to get:
2(2.5) - 1) = 4 which becomes:
5 - 1 = 4 which is also true.
the solution applies to both equations and is confirmed as good.
the point where these lines intersect is (x,y) = (2.5,1).
that intersection point is the solution that is common to both equations.