FreddieBellavia267
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# Which point lies on the graph of the boundary line of the inequality 3y + 4x < 12?

$3 y + 4 x \ \textless \ 12$ Subtract 4 x from both sides $3y+4x-4x \ \textless \ 12-4x$$3y\ \textless \ 12 - 4x$ Divide both sides by 3 : $\frac{3y}{3} \ \textless \ \frac{12-4x}{3}$ $y\ \textless \ \frac{12-4x}{3}$ Image attached hope this helps!