Inequalities and Regions on Graphs

The boundary is a horizontal line, so the inequality is about y.

1. The boundary is the horizontal line y=2.
2. R is above the line, where y is greater than 2, and the line is solid, so y=2 is included.
3. So R is y\ge2. (An inequality such as x\ge2 would give a region to the right of a vertical line.)

A vertical line has an equation x=\ldots; check whether it is solid or dashed.

1. The boundary is the vertical line x=1, and R is to its left, where x is less than 1.
2. The line is dashed, so points on x=1 are not in R: use <, not \le.
3. So R is x<1.

Find the equation of the line in the form y=mx+c first.

1. The line crosses the y-axis at 1 and goes up 1 for every 1 across, so its gradient is 1 and its equation is y=x+1.
2. R is below the line. Check with a shaded point such as (2,0): 0 is less than 2+1=3.
3. The line is solid, so it is included: R is y\le x+1.

Find the equation of the line from where it crosses the axes.

1. The line passes through (0,4) and (4,0): it goes down 1 for every 1 across, so its gradient is -1 and its equation is y=-x+4.
2. R is above the line. Check with a shaded point such as (4,4): 4 is greater than -4+4=0.
3. The line is dashed, so it is not included: R is y>-x+4.

Check each point against every edge of R.

1. R is bounded by the y-axis (x\ge0), the x-axis (y\ge0) and the line through (0,5) and (5,0), which is x+y=5. R is on or below this line, so x+y\le5.
2. (1,2): both coordinates are positive and 1+2=3, which is less than 5, so (1,2) is in R.
3. (3,3) and (2,4) have x+y=6, so they are above the sloping line. (-1,2) has a negative x-coordinate, so it is to the left of the y-axis.

Read the two horizontal lines, and check which one is dashed.

1. The lower boundary is y=1 and is dashed, so y>1.
2. The upper boundary is y=3 and is solid, so y\le3.
3. Combining these gives 1<y\le3.

The missing edge is a vertical line, x=\ldots

1. The edges x=4, y=0 and y=3 are given. The fourth edge is the vertical line x=-1.
2. R is to the right of it, where x is greater than -1, and the line is dashed, so it is not included.
3. The missing inequality is x>-1.

Find the equation y=mx+c of the sloping line, then decide which side of it R is on.

1. The sloping line crosses the y-axis at -1 and goes up 1 for every 1 across, so its equation is y=x-1.
2. R is above this line. Check with the shaded point (0,0): 0 is greater than 0-1=-1.
3. The line is solid, so it is included: y\ge x-1.

Remember that a point on a dashed line is not in R.

1. The dashed sloping line passes through (0,4) and (4,0), so it is x+y=4. R is below it, so x+y<4.
2. (2,2) has x+y=4, so it lies on the dashed line and is not in R.
3. (1,1), (0,2) and (2,0) all have x+y less than 4. The last two are on the solid axes, which are part of R.

Find where the sloping line meets the horizontal line.

1. The corner of R furthest to the right is where the horizontal line y=1 meets the sloping line.
2. The sloping line passes through (0,6) and (6,0), so its equation is x+y=6.
3. On y=1: x+1=6, so x=5. The corner is (5,1).

Find the equation of each edge, then use a point inside R, such as (1,2), to choose each inequality sign.

1. The left and bottom edges are the axes, with R to the right and above: x\ge0 and y\ge0.
2. The top edge is the horizontal line y=4, with R below it: y\le4.
3. The sloping line passes through (1,0) and (2,2), so its gradient is 2 and its equation is y=2x-2.
4. R is above this line: the point (1,2) gives 2\ge2\times1-2=0. So y\ge2x-2. The only option with all four of these inequalities is x\ge0,\ y\ge0,\ y\le4,\ y\ge2x-2.

Read the gradient from two points where the line passes through grid corners.

1. The line passes through (0,2) and (4,4), so its gradient is \frac{4-2}{4-0}=\frac12.
2. Its equation is y=\frac12x+2. Multiply by 2: 2y=x+4.
3. R is below the line and the line is solid, so 2y\le x+4. (The option 2y<x+4 would need a dashed line.)

Find the y-coordinate where the vertical line x=1 meets the bottom edge of R.

1. The bottom edge of R is the sloping line through (-1,0) and (0,1), which is y=x+1.
2. At x=1 this line is at y=2. The line is solid, so (1,2) is in R.
3. Points with x=1 below (1,2) are outside R, so the least possible value is k=2.

Find where the vertical line x=2 enters and leaves R.

1. The bottom edge of R is the solid line y=1.
2. The top edge passes through (0,4) and (6,1), so its gradient is -\frac12 and its equation is y=-\frac12x+4. At x=2, y=-1+4=3.
3. Both edges are solid, so 1\le k\le3. The possible values are 1, 2 and 3.

Substitute the coordinates of P and Q into each inequality.

1. x\ge2: P has x=1, so P is removed. Q has x=2, and 2\ge2 is true, so Q stays.

2. y\ge4 keeps P (as 4\ge4) but removes Q.
3. x+y\ge5 keeps both, because 1+4=5 and 2+3=5. y\le5 also keeps both.
4. So the inequality is x\ge2.

For each sloping line, find its gradient and where it crosses the y-axis, then decide which side of it R is on.

Part (a)

1. The vertical edges are the solid lines x=0 and x=2.
2. R lies between them, and both lines are included, so 0\le x\le2.

Part (b)

1. This line crosses the y-axis at -1 and goes up 1 for every 1 across, so its equation is y=x-1.
2. R is above this solid line, so y\ge x-1.

Part (c)

1. This line crosses the y-axis at 6 and goes down 1 for every 1 across, so its equation is y=-x+6.
2. R is below this dashed line, so y<-x+6.

Work through x=0, x=1, \ldots and count the integer values of y allowed for each.

1. The edges are the y-axis (x\ge0), the line through (0,1) with gradient 1, which is y=x+1, and the line through (0,6) and (3,0), which is y=6-2x.
2. R is above y=x+1 and below y=6-2x. These lines meet where x+1=6-2x, so x=\frac53; there is no part of R to the right of this, so the integer x-coordinates are 0 and 1.
3. At x=0: 1\le y\le6, giving 6 points. At x=1: 2\le y\le4, giving 3 points.

4. Total: 6+3=9 points.

Check each point against every edge; a point on a dashed line is not in R.

1. R is given by x\ge0, y\ge0, y\le-x+4 (solid) and y>x-1 (dashed).
2. A(1,2): 2>1-1=0 and 2\le-1+4=3, so A is in R. D(0,3) is on the solid y-axis, with 3>-1 and 3\le4, so D is in R.
3. B(2,1): x-1=1, so B is on the dashed line and is not in R.
4. C(3,0): x-1=2, and 0 is not greater than 2, so C is below the dashed line and not in R.
5. The two points are A and D.

Imagine the horizontal line y=2 across the grid, and find where it is inside R.

1. The edges are x=0, x=4, y=5 and the sloping line through (0,1) and (4,3), which has gradient \frac12, so it is y=\frac12x+1.
2. R is on or above the sloping line, so for (k,2): 2\ge\frac12k+1, which gives \frac12k\le1, so k\le2.
3. The vertical edges give 0\le k\le4, and y=2 is below the top edge y=5.
4. Both conditions together: 0\le k\le2.

Test a point inside R, such as (3,2), in each inequality.

1. R is to the left of x=4, above y=0 and below y=4, so those three inequalities are correct.
2. The sloping line passes through (0,4) and (2,0), so its gradient is -2 and its equation is y=-2x+4.
3. The shaded point (3,2) gives -2\times3+4=-2, and 2\ge-2, so R is above the line.
4. The last inequality should be y\ge-2x+4.