Solve for x and y:2x+3y=7 (a+b+1)x+(a+2b+2)y=4 (a+b)+1

1.	Solve for x and y:2x+3y=7 (a+b+1)x+(a+2b+2)y=4 (a+b)+1
Answer» If infinite number of solutions,{tex}\\frac{{{a_1}}}{{{a_2}}} = \\frac{{{b_1}}}{{{b_2}}} = \\frac{{{c_1}}}{{{c_2}}}{/tex}or {tex}\\frac{2}{{a + b + 1}} = \\frac{3}{{a + 2b + 2}} = \\frac{7}{{4a + 4b + 1}}{/tex}If\xa0{tex}\\frac{2}{{a + b + 1}} = \\frac{3}{{a + 2b + 2}}{/tex}{tex}\\Rightarrow{/tex} a - b = 1and if {tex}\\frac{3}{{a + 2b + 2}} = \\frac{7}{{4a + 4b + 1}}{/tex}{tex}\\Rightarrow{/tex} 5a - 2b = 11On solving we get,a = 3 and b = 2

Discussion