Pairs such as \(x^{2}y = 18\) and \(xy^{2} = 12\), or \(\dfrac{x}{y} + \dfrac{3y}{x} = 4\) with \(y = x - 2\), can be solved by brute substitution, but there is usually a shorter route: divide one equation by the other, multiply by \(xy\) to clear the fractions, or use \((x + y)^{2} = x^{2} + y^{2} + 2xy\). Whatever the route, the answer is a set of PAIRS, and every pair must satisfy both equations.
Look at the shape of the pair before you start. If both equations are products such as \(xy^{2}\) and \(xy\), DIVIDE one by the other so that a letter cancels (or multiply them to get a power of \(xy\)). If there are fractions such as \(\dfrac{x}{y}\) or \(\dfrac{1}{x}\), multiply EVERY term by \(xy\), noting that \(x \ne 0\) and \(y \ne 0\). If you know \(x^{2} + y^{2}\) and \(xy\), use \((x + y)^{2} = x^{2} + y^{2} + 2xy\) and \((x - y)^{2} = x^{2} + y^{2} - 2xy\). Finish by writing the solutions as pairs and checking each pair in BOTH original equations.
Fig. 1: the circle x² + y² = 25 and the hyperbola xy = 12 meet at four points, (3, 4), (4, 3), (-3, -4) and (-4, -3). Each point is one solution PAIR of the simultaneous equations; (3, -4) lies on the circle but not on the hyperbola
Section 01
Divide one equation by the other.
When both equations are products of powers of \(x\) and \(y\), substitution leads to awkward powers. Dividing one equation by the other cancels a letter at once: \(\dfrac{xy^{2}}{xy} = y\). Multiplying the equations together can also help, because \(x^{2}y \times xy^{2} = (xy)^{3}\). Dividing is safe here because neither side can be zero (a product equal to a non-zero number means \(x \ne 0\) and \(y \ne 0\)). In topic 6 the same idea of combining equations will handle pairs with logarithms.
Worked example Divide to cancel x
Solve the simultaneous equations \(xy^{2} = 12\) and \(xy = 4\).
Neither equation can hold with \(x = 0\) or \(y = 0\), so dividing is allowed.
\(\dfrac{xy^{2}}{xy} = \dfrac{12}{4}\), so \(y = 3\).
Substitute into \(xy = 4\): \(3x = 4\), so \(x = \dfrac{4}{3}\).
Check in the OTHER equation: \(xy^{2} = \dfrac{4}{3} \times 9 = 12\). Correct.
Answer: \(x = \dfrac{4}{3},\ y = 3\)
Worked example Multiply, then divide
Solve \(x^{2}y = 18\) and \(xy^{2} = 12\).
Multiply the equations: \(x^{3}y^{3} = 18 \times 12 = 216\), so \((xy)^{3} = 216\) and \(xy = 6\) (a cube root has only one real value).
\(x^{2}y = x(xy) = 6x = 18\), so \(x = 3\).
\(xy = 6\) then gives \(y = 2\).
Check: \(x^{2}y = 9 \times 2 = 18\) and \(xy^{2} = 3 \times 4 = 12\). Both correct.
Answer: \(x = 3,\ y = 2\)
Your turn
Solve \(xy^{3} = 54\) and \(xy = 6\).
Divide: \(y^{2} = 9\), so \(y = 3\) or \(y = -3\). From \(xy = 6\): \(y = 3\) gives \(x = 2\); \(y = -3\) gives \(x = -2\). Solutions \((2, 3)\) and \((-2, -3)\); both satisfy \(xy^{3} = 54\).
Examiner trap
After dividing, a square such as \(y^{2} = 9\) has TWO values, \(y = 3\) and \(y = -3\). Each gives its own \(x\) from \(xy = 6\), so the answers are \((2, 3)\) and \((-2, -3)\), never the mix \((2, -3)\).
Section 02
Clear the fractions with xy.
Equations with \(x\) and \(y\) in denominators become polynomial equations when you multiply EVERY term, on both sides, by \(xy\). This is only allowed because the original equation needs \(x \ne 0\) and \(y \ne 0\), so state that and reject any solution with \(x = 0\) or \(y = 0\). Then use the other equation, usually by substitution, exactly as in the previous lesson.
\(\dfrac{x}{y} + \dfrac{ky}{x} = c\): multiply by \(xy\) to get \(x^{2} + ky^{2} = cxy\). This often factorises into two linear factors.
\(\dfrac{a}{x} + \dfrac{b}{y} = c\): multiply by \(xy\) to get \(ay + bx = cxy\).
Factorise where you can: each factor gives a linear relation between \(x\) and \(y\) to pair with the other equation.
Worked example Factorise after clearing
Solve \(\dfrac{x}{y} + \dfrac{3y}{x} = 4\) and \(y = x - 2\).
\(x \ne 0,\ y \ne 0\). Multiply every term by \(xy\): \(x^{2} + 3y^{2} = 4xy\).
Rearrange and factorise: \(x^{2} - 4xy + 3y^{2} = (x - y)(x - 3y) = 0\), so \(x = y\) or \(x = 3y\).
\(x = y\) with \(y = x - 2\) gives \(x = x - 2\), which is impossible: no solution from this factor.
\(x = 3y\) with \(y = x - 2\) gives \(y = 3y - 2\), so \(y = 1\) and \(x = 3\).
Solve \(\dfrac{2}{x} + \dfrac{1}{y} = 1\) and \(x + y = 6\).
Multiply by \(xy\): \(2y + x = xy\). With \(x = 6 - y\): \(y + 6 = 6y - y^{2}\), so \(y^{2} - 5y + 6 = 0\), \(y = 2\) or \(3\). Solutions \(x = 4,\ y = 2\) and \(x = 3,\ y = 3\).
Examiner trap
Multiply EVERY term by \(xy\), including the right-hand side. From \(\dfrac{x}{y} + \dfrac{3y}{x} = 4\), writing \(x^{2} + 3y^{2} = 4\) instead of \(x^{2} + 3y^{2} = 4xy\) is the most common lost mark in this kind of question.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Solve \(xy^{2} = 20\) and \(xy = 5\).
Quick check+10 XP
To clear the fractions in \(\dfrac{x}{y} + \dfrac{2y}{x} = 3\), multiply every term by
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Given \(x^{2} + y^{2} = 13\) and \(xy = 6\), find the value of \((x + y)^{2}\).
Quick check+10 XP
For \(x^{2} + y^{2} = 13\) and \(xy = 6\), one solution has \(x = 2\). What is the matching value of \(y\)?
Section 03
Sum of squares and a product.
When you are given \(x^{2} + y^{2}\) and \(xy\), the identities \((x + y)^{2} = x^{2} + y^{2} + 2xy\) and \((x - y)^{2} = x^{2} + y^{2} - 2xy\) turn the pair into two linear equations. Each square root gives \(\pm\), so expect up to four solution pairs (Fig. 1). The alternative is substitution: \(y = \dfrac{12}{x}\) in \(x^{2} + y^{2} = 25\) gives \(x^{4} - 25x^{2} + 144 = 0\), a quadratic in \(x^{2}\).
Worked example Four pairs from two identities
Solve \(x^{2} + y^{2} = 25\) and \(xy = 12\).
\((x + y)^{2} = 25 + 2(12) = 49\), so \(x + y = 7\) or \(x + y = -7\).
\((x - y)^{2} = 25 - 2(12) = 1\), so \(x - y = 1\) or \(x - y = -1\).
\(x + y = 7\) with \(x - y = \pm 1\) gives \((4, 3)\) and \((3, 4)\); \(x + y = -7\) gives \((-3, -4)\) and \((-4, -3)\).
Check by substitution instead: \(x^{4} - 25x^{2} + 144 = (x^{2} - 9)(x^{2} - 16) = 0\), so \(x = \pm 3\) or \(\pm 4\), and \(y = \dfrac{12}{x}\) pairs each one: \(x = 3\) goes with \(y = 4\), not \(y = -4\).
Answer: \((3, 4),\ (4, 3),\ (-3, -4),\ (-4, -3)\)
Worked example A rectangle from its area and diagonal
A rectangle has area 60 cm2 and diagonal 13 cm. Find its length, its width and its perimeter.
Let the sides be \(x\) cm and \(y\) cm with \(x > y > 0\). Then \(xy = 60\) and, by Pythagoras, \(x^{2} + y^{2} = 169\).
\((x + y)^{2} = 169 + 120 = 289\), so \(x + y = 17\) (lengths are positive, so reject \(-17\)).
\((x - y)^{2} = 169 - 120 = 49\), so \(x - y = 7\) (taking \(x > y\)).
Adding: \(2x = 24\), so \(x = 12\) and \(y = 5\). Check: \(12 \times 5 = 60\) and \(144 + 25 = 169\).
Perimeter \(= 2(x + y) = 34\) cm, read straight from \(x + y = 17\).
Answer: 12 cm by 5 cm, perimeter 34 cm
Your turn
Solve \(x^{2} + y^{2} = 20\) and \(xy = 8\).
\((x + y)^{2} = 36\) and \((x - y)^{2} = 4\), so \(x + y = \pm 6\), \(x - y = \pm 2\). Solutions \((4, 2),\ (2, 4),\ (-4, -2),\ (-2, -4)\).
Examiner trap
Do not pair every \(x\) with every \(y\). From \(x = \pm 3\) and \(y = \pm 4\) a careless answer lists \((3, -4)\), but then \(xy = -12\), not \(12\). Use \(y = \dfrac{12}{x}\) (or the signs of \(x + y\) and \(x - y\)) to pair each value, and check every pair in BOTH equations.
Stage 2 · Exam
Exam-style questions
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Solve the simultaneous equations \(x^{2}y = 50\) and \(xy^{2} = 20\).
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Solve \(\dfrac{x}{y} + \dfrac{y}{x} = \dfrac{5}{2}\) and \(x + y = 6\).
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A rectangle has area 24 cm2 and a diagonal of length \(2\sqrt{13}\) cm. Find its perimeter.
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How many solution pairs \((x, y)\) do the equations \(x^{2} + y^{2} = 10\) and \(xy = 3\) have?
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