simultaneous linear inequalities (i don't have money, im only in 7th grade, im sorry)
1 Answer
From the first equation we get
\[2x - 6 + m < 0 \Rightarrow 2x<6-m \Rightarrow x< \frac{6-m}{2}.\]
From the second equation we get
\[4x - m > 0 \Rightarrow 4x>m \Rightarrow x> \frac{m}{4}.\]
Putting these two together, we get
\[ \frac{m}{4} <x< \frac{6-m}{2} \text{or} x\in (\frac{m}{4}, \frac{6-m}{2}). \]

4.8K
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 226 views
- Pro Bono
Related Questions
- Graph the pair of equations in the same rectangular coordinate system: Y=-2x ; y=-2
- Zariski Topology and Regular Functions on Algebraic Varieties in Affine Space
- Algebra 2 problem about a ticket system
- Relating dot product divided with square of the vector while changing basis of vector
- Let $R$ be an integral domain and $M$ a finitely generated $R$-module. Show that $rank(M/Tor(M))$=$rank(M)$
- Need Upper Bound of an Integral
- Prove that $V={(𝑥_1,𝑥_2,⋯,𝑥_n) \in ℝ^n ∣ 𝑥_1+𝑥_2+...+𝑥_{𝑛−1}−2𝑥_𝑛=0}\}$ is a subspace of $\R^n$.
- A word problem about a rectangular carpet