+78 \(Find a linear inequality with the following solution set. Each grid line represents one unit. [asy] size(200); fill((-2,-5)--(5,-5)--(5,5)--(3,5)--cycle,yellow); real ticklen=3; real tickspace=2; real ticklength=0.1cm; real axisarrowsize=0.14cm; pen axispen=black+1.3bp; real vectorarrowsize=0.2cm; real tickdown=-0.5; real tickdownlength=-0.15inch; real tickdownbase=0.3; real wholetickdown=tickdown; void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool useticks=false, bool complexplane=false, bool usegrid=true) { import graph; real i; if(complexplane) { label("$\textnormal{Re}$",(xright,0),SE); label("$\textnormal{Im}$",(0,ytop),NW); } else { label("$x$",(xright+0.4,-0.5)); label("$y$",(-0.5,ytop+0.2)); } ylimits(ybottom,ytop); xlimits( xleft, xright); real[] TicksArrx,TicksArry; for(i=xleft+xstep; i 0.1) { TicksArrx.push(i); } } for(i=ybottom+ystep; i 0.1) { TicksArry.push(i); } } if(usegrid) { xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray(0.1),extend=true),p=invisible);//,above=true); yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.1),extend=true), p=invisible);//,Arrows); } if(useticks) { xequals(0, ymin=ybottom, ymax=ytop, p=black, Ticks("%",TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=black, Ticks("%",TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); } else { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize)); } }; draw((-2,-5)--(3,5),dashed+red, Arrows(size=axisarrowsize)); rr_cartesian_axes(-5,5,-5,5); for( int i = -4; i <= 4; ++i) { draw((i,-5)--(i,5)); draw((-5,i)--(5,i)); } [/asy] (Give your answer in the form $ax+by+c>0$ or $ax+by+c\geq0$ where $a,$ $b,$ and $c$ are integers with no common factor greater than 1.)\)
+2668 I'm not sure if you can, but, believe me, I can't read LaTex or whatever junk that is...
+2668 Does the graph look like this:
Treat the sign as an equal sign for now (we'll worry about it later)
Start by finding the equation in slope-intercept form(\(y = mx + b\)), which can be done with the slope and y-intercept.
The slope from \((x_1, y_1)\) to \((x_2, y_2)\) is \({\text{rise} \over \text{run}} = {y_2 - y_1 \over x_2 - x_1} = {5 - -5 \over 3 - -2} = {10 \over 5} = 2\)
Now, plugging this into m in the slope-intercept equation gives us \(y = 2x + b\)
Next, we need to plug in 1 of the points given to us and solve for b: \(5 = 2 \times 3 + b\), meaning \(b = -1\).
This means that the equation is \(y = 2x - 1\)
Now, simplify as follows:
\(y + 1 = 2x\)
\(y - 2x + 1 = 0\)
\(\color{brown}\boxed{y - 2x + 1 > 0}\)
