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If: 3x-y = 5 then y^2 = (3x_5)^2 => 9x^2 -30x+25 If: 2x^2 + y^2 = 129 then y^2 = 129-2x^2 So: 9x^2 -30x+25 = 129-2x^2 Transposing terms: 11x^2 -30x -104 = 0 Factorizing the above: (11x-52)(x+2) = 0 meaning x = 52/11 or x = -2 By substituting x into the original equation intersections are at: (52/11, 101/11) and (-2, -11)
10%x = x/10. 129/10 = 12.9
129 x .22 = 28.38
2/3 x 129 = 86
If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)