Browse · MATH Print → jmc algebra intermediate Problem Given that 3x+y=10 and x+3y=14, find 10x2+12xy+10y2. Solution — click to reveal Note that 10x2+12xy+10y2=(9x2+6xy+y2)+(x2+6xy+9y2)=(3x+y)2+(x+3y)2=102+142=296. Final answer 296 ← Previous problem Next problem →