Solve $4x + 3 < 6x + 7$ over the real numbers.
A$x > -2$
B$x < -2$
C$x < 2$
D$x > 2$
Answer & Solution
Correct answer: A. $x > -2$
Subtract $6x$: $-2x + 3 < 7$.
Subtract 3: $-2x < 4$.
Divide by $-2$ (negative — **flip** the sign): $x > -2$.
Verify with $x = 0$: $3 < 7$ ✓. With $x = -3$: $-9 < -11$? No. ✓
Related questions
Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE For [[1,2],[2,4]], the determinant equals:If A is invertible, |A^{−1}| equals:For a 3 × 3 non-singular A with |A| = 5, the value of |adj A| is:For a non-singular square matrix A of order n, |adj A| equals:For non-singular A, B of the same order, (AB)^{−1} equals:For a non-singular A, (A^{−1})^{−1} equals: