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√x+y=7. x+√y=11 |
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Answer» first rearrange the equations as followsy - 4 = 3 - {tex}\\sqrt{x}{/tex} ....... i\xa0x - 9 = 2 - {tex} \\sqrt{y}{/tex}\xa0....... iimultiplying i and\xa0ii(y - 4)(x - 9) = (3 - {tex} \\sqrt{x}{/tex})(2 - {tex} \\sqrt{y}{/tex})we can factor LHS({tex} \\sqrt{y}{/tex}\xa0- 2)({tex} \\sqrt{y}{/tex}\xa0+ 2)({tex}\\sqrt{x}{/tex}\xa0+3)({tex} \\sqrt{x}{/tex}\xa0- 3) = (3 - {tex} \\sqrt{x}{/tex})(2 - {tex} \\sqrt{y}{/tex})({tex} \\sqrt{y}{/tex}\xa0- 2)({tex} \\sqrt{x}{/tex}\xa0- 3)({tex} \\sqrt{y}{/tex}\xa0- 2)({tex} \\sqrt{x}{/tex}\xa0- 3) - 1 = 0at least one of the factors must be zero{tex} \\sqrt{y}{/tex}\xa0- 2 = 0 ...... {tex} \\sqrt{y}{/tex}\xa0= 2 ............ y = 4substituting in i3\xa0- {tex} \\sqrt{x}{/tex}\xa0= 0 ........{tex} \\sqrt{x}{/tex}\xa0= 3 ........... x = 9therefore x = 9 and y = 4 first rearrange the equations as followsy - 4 = 3 - sqrt x ....... i\xa0x - 9 = 2 - sqrt y ....... iimultiplying i and\xa0ii(y - 4)(x - 9) = (3 - sqrt x)(2 - sqrt y)we can factor LHS(sqrt y - 2)(sqrt y + 2)(sqrt x +3)(sqrt x - 3) = 3 - sqrt x)(2 - sqrt y)(sqrt y - 2)(sqrt x - 3)(sqrt y - 2)(sqrt x - 3) - 1 = 0at least one of the factors must be zerosqrt y - 2 = 0 ...... sqrt y = 2 ............ y = 4substituting in i3\xa0- sqrt x = 0 ........ sqrt x = 3 ........... x = 9therefore x = 9 and y = 4 |
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