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How many integers satisfy the inequality |10(x+1)/x^2+2x+3|=1? Solution) first thing thats not an inequality,and second thing its very easy if thats the question.the LHS = |10/x^2 + 10/x + 2x + 3| = 110/x^2 will also be an integer only if x^2<10.....i.e. x<3.hence x= 0,+1,-1,+2,-2,+3,-3.but the only value that holds the equality is -1.hence the no. of INTEGERS is only 1.
Find no. of non negative integral solutions x 1 +x 2 +x 3 +4x 4 =20 Solution) 140. Break them into prime factors . Put 4 = 2^2 and every variable will have factors in 2,3,5 with
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Simultaneous equations by substitution: Solve the subsequent simultaneous equations by substitution. 3x + 4y = 6 5x + 3y = -1 Solution: Solve for x: 3x = 6
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