
How do you solve 28 – 7x $\le $ -4 ( -7x - 7 ) ?
Answer
535.8k+ views
Hint: To solve the given inequality, we can first solve the RHS part of the equation which is -4 ( -7x - 7 ) is equal to 28x + 28 . Then we can cancel 28 from both sides. Then we can see that there are no constants in the inequality. The equation will change into 35x > 0, from this we can find the range of x.
Complete step by step solution:
The given inequality is equal to 28 – 7x $\le $ -4 ( -7x - 7 )
Further solving it we can write 28 – 7x $\le $ 28x + 28, subtracting 28 both sides we get -7x $\le $ 28x
Adding both sides 7x we get 35x is greater than equal to 0, $35x\ge 0$ , x is greater than equal to 0 , the value of x is always greater than equal to 0, $x\in \left[ 0,\infty \right)$
Note: If a inequality is in the form kx > 0 , then the range of x is greater than 0 if k is greater than 0 and the range of x is less than 0 if k is less than 0. If k is greater than 0 then $k{{x}^{2}}$ is always greater than equal to 0 no matter what is the value of x. Similarly, $k{{x}^{2}}$ is always less than equal to 0 is k < 0. Always remember while writing infinitely we always use a round bracket because we can never reach infinite value.
Complete step by step solution:
The given inequality is equal to 28 – 7x $\le $ -4 ( -7x - 7 )
Further solving it we can write 28 – 7x $\le $ 28x + 28, subtracting 28 both sides we get -7x $\le $ 28x
Adding both sides 7x we get 35x is greater than equal to 0, $35x\ge 0$ , x is greater than equal to 0 , the value of x is always greater than equal to 0, $x\in \left[ 0,\infty \right)$
Note: If a inequality is in the form kx > 0 , then the range of x is greater than 0 if k is greater than 0 and the range of x is less than 0 if k is less than 0. If k is greater than 0 then $k{{x}^{2}}$ is always greater than equal to 0 no matter what is the value of x. Similarly, $k{{x}^{2}}$ is always less than equal to 0 is k < 0. Always remember while writing infinitely we always use a round bracket because we can never reach infinite value.
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