Solve the following pair of linear equations by the cross multiplication method 8x +5y=9, 3x +2y=9?

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The given pair of linear equations8x + 5y = 9 …(1) …(1)3x + 2y = 4 …(2) …(2)\tby substituting method\tFrom equation (2), 2y = 4 – 3x\t{tex}\\Rightarrow \\;y = \\frac{{4 – 3x}}{2}{/tex}\xa0…(3)\tSubstitute this value of y in equation (1), we get\t{tex}8x + 5\\left( {\\frac{{4 – 3x}}{2}} \\right) = 9{/tex}\t{tex}\\Rightarrow{/tex} 16x + 20 – 15x = 18\t{tex}\\Rightarrow{/tex} x + 20 = 18\t{tex}\\Rightarrow{/tex} x = 18 – 20\t{tex}\\Rightarrow{/tex} x = -2\tsubstituting this value of x in equation (3), we get\t{tex}y = \\frac{{4 – 3( – 2)}}{2} = \\frac{{4 + 6}}{2} = \\frac{{10}}{2} = 5{/tex}\tSo the solution of the given pair of linear equations is x = -2, y = 5.\tby cross-multiplication method\tLet us write the given pair of linear equation is\t8x + 5y – 9 = 0 …(1)\t3x + 2y – 4 = 0 …(2)\tTo solve the equation (1) and (2) by cross multiplication method,\twe draw the diagram below:\t\tThen,\t{tex}\\frac{x}{{(5)( – 4) – (2)( – 9)}} = \\frac{y}{{( – 9)(3) – ( – 4)(8)}}{/tex}{tex} = \\frac{1}{{(8)(2) – (3)(5)}}{/tex}\t{tex}= \\frac{x}{{ – 20 + 18}} = \\frac{y}{{ – 27 + 32}} = \\frac{1}{{16 – 15}}{/tex}\t{tex}\\Rightarrow \\frac{x}{{ – 2}} = \\frac{y}{5} = \\frac{1}{1}{/tex}\t{tex}\\Rightarrow{/tex} x = -2 and y = 5\tHence, the required solution of the given pair of linear equations is x = -2, y = 5.\tVerification : substituting x = -2, y = 5, we find that both the equation (1) and (2) are satisfied as shown below:\t8x + 5y = 8(-2) + 5(5) = -16 + 25 = 9\t3x + 2y = 3(-2) + 2(5) =- 6 + 10 = 4\tHence, the solution is correct.