The solution of the simultaneous equations (1/2)x + (1/3)y = 2 and x+y=1 is:
x = 0, y = 1
x = 1, y = 0
x = 2/3, y = 3/2
x = 10, y = -9
Additional Questions
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If the equation x2-2(k+1)x+(9/2)k=0 has two identical roots then the values of k are: |
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The number which should be subtracted from 5a2-3ab+7b2 to make it equal to a2+ab+b2, is: |
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If x = (1/2) (2p+2q-r), y = (1/3) (-p-2q+3r) and z=(1/5) (3p-4r+5q), then the value of 2x-3y-5z is: |
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The roots of the quadratic equation 6x2-5x+1=0 are: |
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If a = 16, b=25, the value of 1/(a-1/2 - b-1/2) is: |
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3a2 (ab+bc+ca) = |
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x4y-xy4 = |
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Factors of 6a2-25a+4 are: |
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The correct relationship after eliminating x, y and z from x+y = a, y+z=b and z+x = c and x+y+z = m, is: |
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If r = at2 and s = 2at, the relation among s, r and a is: |
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