Question 1 (i)
Solve the following pairs of equations by reducing them to a pair of linear equations:
12x+13y=2
13x+12y=136
12x+13y=2
13x+12y=136
Let 1x=p and 1y=q, then the equations changes as below:
p2+q3=2
⇒3p+2q−12=0 ... (i)
p3+q2=136
⇒2p+3q−13=0 ... (ii)
Compare the given equations with
a1x+b1y+c1=0 and a2x+b2y+c2=0.
By cross-multiplication method, we get
xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1
p−26−(−36)=q−24−(−39)=19−4
p10=q15=15
p10=15 and q15=15
p=2 and q=3
1x=2 and 1y=3
Hence, x=12 and y=13