reduce( randRangeNonZero( -5, 5 ), randRange( 1, 5 ) ) randRangeNonZero( max( -10, -10 - SLOPE_FRAC[0] ), min( 10, 10 - SLOPE_FRAC[0] ) ) SLOPE_FRAC[0] / SLOPE_FRAC[1] SLOPE === 1 ? "" : ( SLOPE === -1 ? "-" : fraction( SLOPE_FRAC[0], SLOPE_FRAC[1], true, true ) ) randRangeNonZero( -3, 3 ) SLOPE_FRAC[0] * -MULT SLOPE_FRAC[1] * MULT SLOPE_FRAC[1] * YINT * MULT randFromArray([ true, false ]) randFromArray([ "<", ">", "≤", "≥" ]) B < 0 ? { "<": ">", ">": "<", "≤": "≥", "≥": "≤" }[ COMP ] : COMP COMP === "<" || COMP === "≤" COMP === "≥" || COMP === "≤" SLOPE_FRAC[1] YINT + SLOPE_FRAC[0] -SLOPE_FRAC[1] YINT - SLOPE_FRAC[0] (YINT < 0 && !LESS_THAN) || (YINT > 0 && LESS_THAN)

What is the inequality represented by this graph?

graphInit({ range: 11, scale: 20, axisArrows: "<->", tickStep: 1, labelStep: 1, gridOpacity: 0.05, axisOpacity: 0.2, tickOpacity: 0.4, labelOpacity: 0.5 }); label( [ 0, -11 ], "y", "below" ); label( [ 11, 0 ], "x", "right" ); var dash = INCLUSIVE ? "" : "- "; style({ stroke: BLUE, strokeWidth: 2, strokeDasharray: dash }, function() { line( [ -11, -11 * SLOPE + YINT ], [ 11, 11 * SLOPE + YINT ] ).toBack(); }); graph.shadeEdge = (LESS_THAN ? 11 < YINT : 11 > YINT) ? 11: -11; style({ fill: BLUE, stroke: null, opacity: KhanUtil.FILL_OPACITY }, function() { graph.shading = path([ [ 11, graph.shadeEdge ], [ 11, 11 * SLOPE + YINT ], [ -11, -11 * SLOPE + YINT ], [ -11, graph.shadeEdge ] ]); });

`y`COMP SLOPE`\space x + ` YINT

To find the equation of a linear inequality you should first find the equation of the line that forms the boundary of the solution set. This line is shown on the graph above.

One way to find the equation of this line is to choose two points on the line and find the slope and y-intercept from there. Two points on this line are `(X1,Y1)` and `(X2,Y2)`.

Substitute both points into the equation for the slope of a line.

`m = \dfrac{Y2 - negParens(Y1)}{X2 - negParens(X1)} = fractionReduce( Y2 - Y1, X2 - X1 )`

To find `b`, we can substitute in either of the two points into the equation with solved slope.

Using the first point `(X1, Y1)`, substitute `y = Y1` and `x = X1`:

`Y1 = (fractionReduce( Y2 - Y1, X2 - X1 ))(X1) + b`

`b = Y1 - fractionReduce( X1 * ( Y2 - Y1 ), X2 - X1 ) = fractionReduce( Y1 * (X2 - X1) - X1 * ( Y2 - Y1 ), X2 - X1 )`

The equation of the line is `y = ( SLOPE === -1 ? "-" : ( SLOPE === 1 ? "" : fractionReduce( Y2 - Y1, X2 - X1 ))) x + fractionReduce( Y1 * (X2 - X1) - X1 * ( Y2 - Y1 ), X2 - X1 )` (the value of `m` is `SLOPE`).

Now that we have the equation for the boundary line, we need to decide which inequality sign to use.

If we pick a point on the line, let's say `(X1,Y1)`, we can see that points LESS_THAN ? "below" : "above" that point are all shaded in. These are the points where `x = X1 ` but `y` is LESS_THAN ? "less than" : "greater than" Y1 . So we should use a LESS_THAN ? "< or ≤" : "> or ≥" sign.

Another way to see this is to try plugging in a point. The easiest point to plug in is `(0,0)`:

`y \; ? \; ( SLOPE === -1 ? "-" : ( SLOPE === 1 ? "" : fractionReduce( Y2 - Y1, X2 - X1 ))) x + fractionReduce( Y1 * (X2 - X1) - X1 * ( Y2 - Y1 ), X2 - X1 )`

`0 \; ? \; ( SLOPE === -1 ? "-1*" : ( SLOPE === 1 ? "" : fractionReduce( Y2 - Y1, X2 - X1 ) + "*")) 0 + fractionReduce( Y1 * (X2 - X1) - X1 * ( Y2 - Y1 ), X2 - X1 )`

Since `(0,0)` is ORI_IN ? "" : "not" in the shaded area, this expression must be ORI_IN ? "true" : "false" So, the ? must be either LESS_THAN ? "< or ≤" : "> or ≥" .

The line of the graph is INCLUSIVE ? "solid" : "dashed", so the points on the boundary are INCLUSIVE ? "" : "not" in the set of solutions for this inequality.

So, we choose the COMP sign, and the final inequality is `y COMP PRETTY_SLOPE x + YINT`.