One, none, or infinitely many
Tonight's idea
An equation is a balance. Whatever you do to one pan you do to the other. With \(x\) on both sides there is one new move: take the same number of \(x\)'s off each pan.
Try it on \(3x + 5 = 3x + 7\). Take \(3x\) off each side: \(5 = 7\). False, whatever \(x\) is. No weight levels the pans, so the equation has no solution.
Now \(2(x + 1) = 2x + 2\). Expand the left: \(2x + 2 = 2x + 2\). Take \(2x\) off: \(2 = 2\). True, whatever \(x\) is. Every number is a solution. Such an equation is called an identity.
Collect the \(x\)'s, then, and one of three things is left.
\(x = 5\): one solution. \(5 = 7\): no solution. \(2 = 2\): every number is a solution.
You can tell which before you finish. Expand both sides and compare the coefficients of \(x\), the numbers multiplying it. Different coefficients: the \(x\)'s survive, and there is exactly one solution. Equal coefficients: the \(x\)'s vanish and the constants decide. Equal constants means every \(x\); unequal means none.
Watch one: \(3(2x - 1) = 4x + 7\), then its twin
Expand: \(6x - 3 = 4x + 7\). The coefficients, 6 and 4, differ, so expect one solution. Take \(4x\) off both sides: \(2x - 3 = 7\). Add 3: \(2x = 10\), so \(x = 5\). Check: \(3(9) = 27\) and \(20 + 7 = 27\).
Now change the \(4x\) to \(6x\): \(3(2x - 1) = 6x + 7\). Expanding gives \(6x - 3 = 6x + 7\). The coefficients match, so the \(x\)'s will cancel, leaving \(-3 = 7\). False. No solution, and you knew it after one line.
Your turn
- Solve, and check your answer in the original equation: (a) \(7x - 4 = 3x + 12\); (b) \(2(x + 3) = 5x - 9\).
- Say how many solutions each has, without solving all the way: (a) \(4x + 9 = 4x - 1\); (b) \(3(x - 2) = 3x - 6\); (c) \(2x + 8 = 8 + 3x\).
- Solve: (a) \(5 - 2(x - 3) = 3(x + 1) - 4x\); (b) \(\dfrac{x}{2} + 3 = \dfrac{x}{3} + 5\). For (b), multiply every term by 6 first.
- Consider \(kx + 6 = 2(x + 3) + 5x\), where \(k\) is a fixed number. (a) For which \(k\) does it have infinitely many solutions? (b) Explain why no value of \(k\) makes it have no solution at all. (c) Change the 6 on the left to 5. Now which \(k\) gives no solution?
Stretch: a letter in the coefficient
Solve \(a(ax + 1) = 4x + 2\) for \(x\), treating \(a\) as a known number. For which values of \(a\) is there no solution? Infinitely many? For every other \(a\), simplify your answer as far as it goes. It helps that \(a^2 - 4 = (a - 2)(a + 2)\); check this by expanding.
Answers
- (a) \(4x = 16\), \(x = 4\). Check: \(24 = 24\). (b) \(2x + 6 = 5x - 9\), so \(15 = 3x\), \(x = 5\). Check: \(16 = 16\).
- (a) Coefficients match, constants differ: none. (b) Expand: \(3x - 6 = 3x - 6\), an identity: every \(x\). (c) Coefficients 2 and 3 differ, so exactly one: subtracting \(2x + 8\) from both sides gives \(0 = x\). Zero is a perfectly good solution.
- (a) \(11 - 2x = 3 - x\), so \(x = 8\). (b) \(3x + 18 = 2x + 30\), so \(x = 12\).
- The right side is \(7x + 6\). (a) \(k = 7\) gives \(7x + 6 = 7x + 6\): every \(x\). (b) Both constants are 6. When the \(x\)'s cancel, what is left is \(6 = 6\), never false; when they do not cancel there is one solution, \(x = 0\). (c) \(kx + 5 = 7x + 6\): with \(k = 7\) the \(x\)'s cancel and leave \(5 = 6\). No solution.
Stretch. \(a^2 x + a = 4x + 2\), so \((a^2 - 4)x = 2 - a\). If \(a = 2\): \(0 = 0\), every \(x\). If \(a = -2\): \(0 = 4\), no solution. Otherwise \(x = \dfrac{2 - a}{(a - 2)(a + 2)} = \dfrac{-1}{a + 2}\). Check with \(a = 0\): the equation is \(0 = 4x + 2\), so \(x = -\tfrac12\), and the formula gives \(-\tfrac12\).