Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x+5y=9; 3x+2y=4

Substitution method:
8x+5y=9 ∴y=9−8x/5 (1)
3x+2y=4.. (2)
Substitute eq(1) in eq (2), we get
3x+2(9−8x/5)=4
Multiply by 5 on both sides, we get
15x+18−16x=20 ∴ −x=2 ∴ x=−2
Substitute x=−2 in eq(1), we get
y=9−8(−2)/5 ∴y=25/5=5
So,(x,y)=(−2,5)
Cross-multiplication method:
8x+5y=9.. (1)
So,a1=8,b1=5,c1=−9
3x+2y=4.. (2)
So,a2=3,b2=2,c2=−4
∴x=(b1c2−b2c1)/(a1b2−a2b1) ∴x=(5(−4)−2(−9))/(8(2)−3(5)) = −20+18/16−15=−2/1
∴y=(c1a2−c2a1)/(a1b2−a2b1) ∴y=(−9(3)−(−4)8)/(8(2)−3(5)) =−27+32/16−15=5/1
So,(x,y)=(−2,5)