If x=2, then 7x = 14 and if y=5 then 4y=20.
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7x+4y = 39 2x+4y = 14 Subtract the equations in order to eliminate y: 5x = 25 Divide both sides by 5 to find the value of x: x = 5 Substitute the value of x into the original equations to find the value of y: Therefore: x = 5 and y = 1
7x + 5y 7(3) + 5(2) 21 + 10 = 31
(5, 2)
If: x -y = 2 then x^2 = y^2 +4y +4 If: x^2 -4y^2 = 5 then x^2 = 4y^2 +5 So: 4y^2 +5 = y^2 +4y +4 Transposing terms: 3y^2 -4y +1 = 0 Factorizing the above: (3y -1)(y -1) = 0 meaning y = 1/3 or y = 1 Substitution into the original linear equation intersections are at: (7/3, 1/3) and (3, 1)
2log4(7x) - 5 = 0 2log4(7x) = 5 log4(7x) = 5/2 7x = 45/2 =(41/2)5 = 25 = 32 x = 32/7