The first equation gives \(y\) on its own, so put that expression in place of the \(y\) in the second one.

\(x(x - 3) = 28\)

\(x^2 - 3x = 28\)

A quadratic can only be factorised once one side is zero:

\(x^2 - 3x - 28 = 0\)

\((x + 4)(x - 7) = 0\)

\(x = -4\) or \(x = 7\)

Then \(y = x - 3\) for each:

when \(x = -4\), \(y = -7\); when \(x = 7\), \(y = 4\).

Both pairs really do work in the second equation: \((-4) \times (-7) = 28\) and \(7 \times 4 = 28\).

The first equation has both letters squared, so there is no rearranging it into anything useful. The second one already gives \(y\) on its own, so that is the one to substitute.

Replace every \(y\) in the first equation with \(-2x - 10\):

\(x^2 + (-2x - 10)^2 = 65\)

Expand the squared bracket — remembering it has three terms, not two:

\(x^2 + 4x^2 + 40x + 100 = 65\)

\(5x^2 + 40x + 35 = 0\)

Every term divides by \(5\), and taking that factor out is what turns this into something factorisable:

\(x^2 + 8x + 7 = 0\)

\((x + 7)(x + 1) = 0\)

\(x = -7\) or \(x = -1\)

Now find each \(y\) from the straight-line equation, which is much less work than the circle and cannot give a spurious extra answer:

when \(x = -7\), \(y = -2(-7) - 10 = 4\)

when \(x = -1\), \(y = -2(-1) - 10 = -8\)

So the line cuts the circle at \((-7, 4)\) and \((-1, -8)\). Pair the answers up: an \(x\) from one solution with the \(y\) from the other is not a point on either graph.

Both equations already say what \(y\) is, so the quickest route is to set the two right-hand sides equal to each other.

\(x^2 + 5x - 8 = 2x - 4\)

Bring everything to the left so that one side is zero:

\(x^2 + 3x - 4 = 0\)

\((x + 4)(x - 1) = 0\)

\(x = -4\) or \(x = 1\)

Put each value back into the straight line — the easier of the two equations, and a good habit because the curve would need squaring again:

when \(x = -4\), \(y = 2(-4) - 4 = -8 - 4 = -12\)

when \(x = 1\), \(y = 2(1) - 4 = 2 - 4 = -2\)

The line crosses the curve at \((-4, -12)\) and \((1, -2)\).

Give the two numbers letters and write down what you are told.

\(x + y = 8\)   (1)

\(x^2 + y^2 = 40\)   (2)

Equation (1) is the linear one, so make \(y\) the subject of it and substitute into (2).

\(y = 8 - x\)

\(x^2 + (8 - x)^2 = 40\)

\(x^2 + 64 - 16x + x^2 = 40\)

\(2x^2 - 16x + 24 = 0\)

Every term is even, so halve the whole equation before factorising:

\(x^2 - 8x + 12 = 0\)

\((x - 2)(x - 6) = 0\)

\(x = 2\) or \(x = 6\)

The two answers are not two different situations here — they are the same pair of numbers taken in the two possible orders. If \(x = 2\) then \(y = 8 - 2 = 6\), and the other way round gives the same pair back.

The two numbers are \(2\) and \(6\).

Call the width \(x\) cm.

The perimeter counts each side twice, so the length and the width add up to half of it:

\(48 \div 2 = 24\)

That makes the length \((24 - x)\) cm, and the area gives a second equation:

\(x(24 - x) = 119\)

Multiply out and collect everything on one side. Writing it with a positive \(x^2\) term is worth the extra line:

\(24x - x^2 = 119\)

\(x^2 - 24x + 119 = 0\)

\((x - 7)(x - 17) = 0\)

\(x = 7\) or \(x = 17\)

Both roots are real answers, because they are the two sides: a width of \(7\) cm goes with a length of \(17\) cm, and a width of \(17\) cm is the same rectangle turned round.

The length is \(17\) cm and the width is \(7\) cm.

Both equations give \(y\), so the two right-hand sides are equal wherever the graphs meet:

\(x^2 + 6x - 20 = 4x - 5\)

Take everything to the left:

\(x^2 + 2x - 15 = 0\)

Factorise it. The two solutions multiply to \(-15\) and add to \(-2\), and the question has already handed over one of them, \(x = -5\), so \((x + 5)\) has to be one of the two factors:

\((x + 5)(x - 3) = 0\)

\(x = -5\) or \(x = 3\)

The point given uses \(x = -5\), so the other one has \(x = 3\).

Put that into the straight line, which is the easier equation:

\(y = 4 \times 3 - 5 = 7\)

The other point is \((3, 7)\).

"Meet" means the same \(x\) and the same \(y\) satisfy both equations, so set the two right-hand sides equal:

\(x^2 +x - 8 = 7x - 17\)

\(x^2 - 6x + 9 = 0\)

That quadratic is a perfect square:

\((x - 3)^2 = 0\)

A squared bracket is zero only when the bracket itself is zero, and there is only one way for that to happen:

\(x = 3\)

One repeated solution rather than two different ones is exactly what "meet at exactly one point" means — the line is a tangent to the curve, touching it instead of cutting through.

\(y = 7(3) - 17 = 21 - 17 = 4\)

The point is \((3, 4)\).

Find the two points first, then put Pythagoras on the gap between them.

Both equations give \(y\), so set the right-hand sides equal:

\(x^2 - 3 = 3x + 7\)

\(x^2 - 3x - 10 = 0\)

\((x + 2)(x - 5) = 0\)

\(x = -2\) or \(x = 5\)

Put each back into the line to get its \(y\):

\(A = (-2, 1)\) and \(B = (5, 22)\)

Now the distance between them. The horizontal gap is \(7\) and the vertical gap is \(21\), so

\(AB^2 = 7^2 + 21^2 = 49 + 441 = 490\)

\(AB = \sqrt{ 490 }\)

\(490 = 49 \times 10\), and \(49\) is a square number, so its root comes outside and \(10\) stays under:

\(AB = 7\sqrt{ 10 }\)

The answer is always a surd here, never a whole number, and that is not an accident: the length works out as \(\text{gap} \times \sqrt{1 + m^2}\), and \(1 + m^2\) is a square number only when the gradient is zero.

Two graphs meet where a pair \((x, y)\) satisfies both equations at once, so substitute the line into the circle and see whether the resulting equation has any solutions.

\(x^2 + (-3x + 31)^2 = 64\)

Expand the squared bracket and collect:

\(x^2 + 9x^2 - 186x + 961 = 64\)

\(10x^2 - 186x + 897 = 0\)

Now the discriminant, which decides how many solutions that has:

\(b^2 - 4ac = 34596 - 35880 = -1284\)

\(-1284\) is negative. The quadratic formula would need the square root of a negative number, and there is no such real number, so the equation has no solutions.

No value of \(x\) puts a point on both graphs, so the line and the circle do not meet.

A tangent touches the curve at exactly one point, so setting the two expressions for \(y\) equal has to give an equation with exactly one solution.

\(x^2 - 12x + 8 = -2x + c\)

\(x^2 - 10x + 8 - c = 0\)

An equation has exactly one solution when its discriminant is zero. Here \(a = 1\), \(b = -10\) and the constant is \(8 - c\):

\((-10)^2 - 4(8 - c) = 0\)

\(4 \times 25 - 4(8 - c) = 0\)

Divide through by \(4\):

\(25 - 8 + c = 0\)

\(c = 8 - 25 = -17\)

Check: with that value the equation becomes \((x - 5)^2 = 0\), which has the single solution \(x = 5\) — the point where the line touches.