We have two equations,
15(x−2)=14(1−y)
and
26x+3y+4=0
If we solve it using cross-multiplication, we use the following arrangement, after converting equations to the standard form.
What are the values of A and B?
A = 5, B = 3
15(x−2)=14(1−y)
4(x–2)=5(1–y)
4x–8–5+5y=0
4x+5y–13=0−−−−−−(1)
26x+3y+4=0−−−−−−(2)
Using cross multiplication method
Hence A = 5 and B = 3.
x20+39=y(−13)(26)−16=112−130
x=59−118=−12
y=−354−118=3