Put the \(x\)-coordinate of the point into the right-hand side and see what it comes to.

\(-3 \times (-5) + 2 = 17\)

The point has \(y = 19\), and \(19\) is greater than \(17\), so the statement \(19 \geq 17\) is true.

Answer: Yes

A point on the line has \(y = 3x - 8\) exactly.

The inequality is \(\gt\), which does not allow the two sides to be equal, so the points on the line are not part of the region.

The boundary line is drawn dashed.

The boundary is the vertical line through \(2\) on the \(x\)-axis.

The inequality is \(\lt\), which does not allow \(x\) to equal \(2\), so the points on that vertical line are not part of the region.

The boundary line is drawn dashed.

For any point on the line, \(y\) is exactly \(2x + 5\).

A point below the line has the same \(x\) but a \(y\) that is lower down the grid than the value the line gives. The points on the line itself are not wanted, so the inequality is strict.

\(y \lt 2x + 5\)

This time the points on the line are wanted as well as the ones higher up the grid than it, so the inequality allows the two sides to be equal.

\(y \geq 2x + 5\)

Test each point against all three inequalities. A point only counts if it passes every one of them, so one failure is enough to rule it out.

\((4, 1)\) fails \(x \gt 5\), because \(4\) is not more than \(5\).

\((6, 4)\) fails \(y \lt 4\), because \(4\) is not less than \(4\).

\((8, 3)\) fails \(x + y \lt 11\), because \(8 + 3 = 11\).

\((6, 1)\) passes all three: \(6 \gt 5\), \(1 \lt 4\), and \(6 + 1 = 7\), which is less than \(11\).

Answer: \((6, 1)\)

Take the two boundaries one at a time.

The vertical line is \(x = -2\). Every point to the left of it has an \(x\)-coordinate less than \(-2\), so the first inequality is \(x \lt -2\).

The sloped line is \(y = -x - 1\). A point above it has the same \(x\) but a \(y\) that is greater than the value the line gives, so the second inequality is \(y \gt -x - 1\).

Neither line is part of \(R\), so both inequalities are strict — no \(\leq\) or \(\geq\).

\(x \lt -2\) and \(y \gt -x - 1\)

Work along the possible values of \(x\), smallest first.

When \(x = 1\), the third inequality gives \(y \leq 3 - 1 = 2\), and the second gives \(y \geq -1\). That is \(4\) whole numbers.

Every time \(x\) goes up by one there is one fewer value of \(y\) left, so the counts are \(4\), then \(3\), and so on down to \(1\).

Adding them up gives \(10\) points.

Pushing \(x\) as far right as possible leaves \(y\) as small as it can be, and the smallest \(y\) is allowed to be is \(-1\).

Put \(y = -1\) into \(x + y \leq 3\):

\(x \leq 3 - (-1) = 4\)

So the point is \((4, -1)\).

Substitute \(x = 0\) and \(y = 3\) into each inequality.

(1) \(0 \gt 0\) is false

(2) \(3 \gt -2\) is true

(3) \(2x + y = 3\) when \(x = 0\) and \(y = 3\), so \(3 \leq 4\) is true

The point is on the edge given by inequality (1), where the two sides are equal, and that inequality does not allow equality. The point does not satisfy all three inequalities.

Answer: No.

Substitute \(x = 2\) and \(y = -2\) into each inequality.

(1) \(2 \gt 0\) is true

(2) \(-2 \gt -2\) is false

(3) \(2x + y = 2\) when \(x = 2\) and \(y = -2\), so \(2 \leq 4\) is true

The point is on the edge given by inequality (2), where the two sides are equal, and that inequality does not allow equality. The point does not satisfy all three inequalities.

Answer: Yes.

The boundary of the region is the line \(y = -2x + 1\). It crosses the \(y\)-axis at \((0, 1)\) and passes through \((1, -1)\). The inequality allows "equal to", so the line is part of the region and is drawn solid.

To choose the side, test a point that is not on the line. Take \((0, 2)\): its \(y\)-coordinate is \(2\), and on the line at \(x = 0\) the value is \(1\).

\(2 ≥ 1\) is true, so \((0, 2)\) is in the region. It is above the line, and so is the whole region.

The region below the line is the one that is not wanted.

Check each line of the graph against its own inequality, three things at a time: the style of the line, the side that is shaded, and where the line crosses the axis.

First inequality, \(y ≤ x - 2\). The boundary is \(y = x - 2\), which crosses the \(y\)-axis at \(-2\). Joe's line crosses at \(-2\): correct. The sign includes equals, which keeps the points on the line, so the line should be solid; Joe drew it dashed: a fault. The sign asks for the points below the line; the shading is below Joe's line: correct.

Second inequality, \(y ≤ -2 - x\). The boundary is \(y = -2 - x\), which crosses the \(y\)-axis at \(-2\). Joe's line crosses at \(-2\): correct. The sign includes equals, which keeps the points on the line, so the line should be solid; Joe drew it solid: correct. The sign asks for the points below the line; the shading is above Joe's line: a fault.

Third inequality, \(x < 3\). The boundary is the vertical line \(x = 3\). Joe's vertical line is at \(x = 3\): correct. The sign is strict, which leaves out the points on the line, so the line should be dashed; Joe drew it dashed: correct. The sign asks for the points to the left of the line; the shading is to the left of Joe's line: correct.

Only two lines have a fault, so the two criticisms are:

(1) The line for the first inequality is drawn dashed, but it should be solid because the inequality includes equals.

(2) The shading is on the wrong side of the line for the second inequality: the region should be below the line.

Take the three lines that are parallel to an axis first. Each one is a straight read from the grid.

The region lies to the right of the line \(x = -4\), so \(x \geq -4\).

The region lies above the line \(y = -2\), so \(y \geq -2\).

The region lies below the line \(y = 4\), so \(y \leq 4\).

The fourth line slopes and is not labelled, so find its equation from two points on it. Two points on the line are \((-1, -2)\) and \((1, 4)\).

Going from \((-1, -2)\) to \((1, 4)\), \(y\) changes by \(6\) while \(x\) changes by \(2\), so the gradient is \(\frac{ 6 }{ 2 } = 3\).

The line is \(y = 3x + c\). It passes through \((-1, -2)\), so \(-2 = -3 + c\) and \(c = -2 + 3 = 1\).

So the line is \(y = 3x + 1\). To find which side of it \(R\) is on, test the corner \((-4, -2)\) of the region: the line gives \(y = -11\) at \(x = -4\), and \(-2\) is greater than \(-11\), so \(R\) is above the line and \(y \geq 3x + 1\).

The four inequalities are \(x \geq -4\), \(y \geq -2\), \(y \leq 4\) and \(y \geq 3x + 1\).