QuanticaPrealgebraOpen in the courseIn 6.2 the unknown sat on one side and the plain numbers on the other. Real situations are not always so tidy, and the same unknown can show up on both sides at once. Picture two scooter apps, one costing \(2m + 5\) and the other \(3m\). They cost the same when \(2m + 5 = 3m\), which has \(m\) on both pans. You cannot isolate the variable while copies sit on each side, so the one new move is to subtract a variable term from both sides until the unknown is on one side only. Everything after that is the same work you did in 6.2.
Problem
A locker rental costs \(7x + 2\) and a bike rental costs \(4x + 17\), each over the same number of days \(x\), and today they came out equal, so \(7x + 2 = 4x + 17\). The new idea is that the balance rule lets you take the same term off both pans, and that term can be a variable. Subtract \(4x\) from both sides to clear the \(x\) off the right, then solve the one-sided equation. Solve \(7x + 2 = 4x + 17\) and type the value of \(x\).
Show a hint
- Both sides carry an \(x\) term. Subtract the smaller one, \(4x\), from each side so the variable survives on only one side.
- After subtracting \(4x\) you get \(3x + 2 = 17\). Now it is an ordinary two-step equation, subtract 2 then divide by 3.
Show the full solution
Subtract \(4x\) from both sides. On the left \(7x - 4x = 3x\), and the right loses its \(x\) term, leaving \(3x + 2 = 17\). Subtract 2 to get \(3x = 15\), then divide by 3 for \(x = 5\). Both sides come out to 37, so \(\boxed{5}\). Taking the same variable term off both pans is legal for the same reason taking off the same number is, the scale stays level.
Problem
When the bigger pile of variables is on the right, a little planning avoids negatives. In \(9 + 3x = 7x + 1\), subtracting \(7x\) would leave a negative \(x\) term, so instead subtract the smaller variable term, \(3x\), from both sides so the survivor stays positive. Solve \(9 + 3x = 7x + 1\) and type the value of \(x\).
Show a hint
- Compare the two variable terms, \(3x\) and \(7x\). Subtract the smaller, \(3x\), from both sides to keep the remaining \(x\) term positive.
- After subtracting \(3x\) you get \(9 = 4x + 1\). Subtract 1, then divide by 4.
Show the full solution
The variable terms are \(3x\) and \(7x\). Subtract the smaller, \(3x\), from both sides so the survivor stays positive. The left becomes \(9\) and the right becomes \(4x + 1\), giving \(9 = 4x + 1\). Subtract 1 from both sides to get \(8 = 4x\), then divide by 4 to get \(x = 2\). A quick check, the original left is \(9 + 3(2) = 15\) and the right is \(7(2) + 1 = 15\), so \(\boxed{2}\).
Problem
Before moving anything across, tidy each side on its own. The left side of \(5x + 4 - 2x = x + 12\) has like terms hiding in it, so combine them first, then gather the variable onto one side. Solve \(5x + 4 - 2x = x + 12\) and type the value of \(x\).
Show a hint
- The left side is not in simplest form. Combine \(5x\) and \(-2x\) into \(3x\), so the equation reads \(3x + 4 = x + 12\).
- Now subtract \(x\) from both sides to get \(2x + 4 = 12\), then finish with the usual two steps.
Show the full solution
Combine the like terms on the left, \(5x - 2x = 3x\), so the equation is \(3x + 4 = x + 12\). Subtract \(x\) from both sides to get \(2x + 4 = 12\), subtract 4 to get \(2x = 8\), then divide by 2 for \(x = 4\). Both sides come out to 16, so \(\boxed{4}\). Tidying each side on its own first keeps the gathering step from getting messy.
Problem
A group can hide the variable until you distribute. In \(2(3x - 4) = 4x + 6\), expand the left side with the distributive property from 6.1, then collect the variable on one side. Solve \(2(3x - 4) = 4x + 6\) and type the value of \(x\).
Show a hint
- Distribute the 2 across the group first. \(2(3x - 4)\) becomes \(6x - 8\), so the equation is \(6x - 8 = 4x + 6\).
- Subtract \(4x\) from both sides to get \(2x - 8 = 6\), then add 8 and divide by 2.
Show the full solution
Distribute the 2 on the left, \(2(3x - 4) = 6x - 8\), so the equation becomes \(6x - 8 = 4x + 6\). Gather the variable by subtracting \(4x\) from both sides, leaving \(2x - 8 = 6\). Add 8 to both sides to get \(2x = 14\), then divide by 2 to get \(x = 7\). Check, the left is \(2(3 \cdot 7 - 4) = 2(17) = 34\) and the right is \(4(7) + 6 = 34\), so \(\boxed{7}\).
Problem
Nothing says the meeting value has to be a whole number. Solve \(8x - 3 = 3x + 4\), gathering the variable onto one side, and let the answer be a fraction if that is where it lands. Type \(x\) as a fraction in lowest terms.
Show a hint
- Subtract \(3x\) from both sides to get \(5x - 3 = 4\).
- Add 3 to get \(5x = 7\), then divide both sides by 5. The answer is a fraction, so leave it in lowest terms.
Show the full solution
Subtract \(3x\) from both sides to get \(5x - 3 = 4\). Add 3 to get \(5x = 7\), then divide by 5 for \(x = \tfrac{7}{5}\), which is already lowest terms since 7 and 5 share no common factor. Both sides come out to \(\tfrac{41}{5}\), so \(\boxed{7/5}\). A fraction answer is an ordinary result here, nothing needs rounding or fixing.
Problem
When a fraction sits in the equation, clear it first. In \(\dfrac{x}{3} + 4 = x - 2\), multiply every term on both sides by \(3\) to wipe out the denominator, then solve the both-sides equation that remains. Solve \(\dfrac{x}{3} + 4 = x - 2\) and type the value of \(x\).
Show a hint
- Multiply every term on both sides by 3 to clear the denominator. \(\tfrac{x}{3}\) becomes \(x\), the 4 becomes 12, and the right side \(x - 2\) becomes \(3x - 6\).
- Now solve \(x + 12 = 3x - 6\). Subtract \(x\), then finish.
Show the full solution
Clear the fraction by multiplying every term on both sides by 3. The left \(\tfrac{x}{3} + 4\) becomes \(x + 12\), and the right \(x - 2\) becomes \(3x - 6\), so the equation is \(x + 12 = 3x - 6\). Subtract \(x\) from both sides to get \(12 = 2x - 6\), add 6 to get \(18 = 2x\), then divide by 2 to get \(x = 9\). Check in the original, the left is \(\tfrac{9}{3} + 4 = 3 + 4 = 7\) and the right is \(9 - 2 = 7\), so \(\boxed{9}\).
Problem
Try to solve \(6x + 5 = 6x - 2\). Gather the variable terms by subtracting \(6x\) from both sides and watch what remains. How many solutions does this equation have? Type the single number.
Show a hint
- Subtract \(6x\) from both sides and watch the variable disappear entirely.
- You are left with \(5 = -2\), which is never true. An equation that reduces to a false statement has no solution, so the count is 0.
Show the full solution
Gather the variable by subtracting \(6x\) from both sides. Every \(x\) vanishes, and you are left with \(5 = -2\). That statement is false, and no value of \(x\) can ever make \(6x + 5\) equal \(6x - 2\), since the two sides always differ by 7. So the equation has no solution, a count of \(\boxed{0}\).
Problem
Now try \(3(2x + 4) = 6x + 12\). Distribute the left side, gather the variable, and notice what is left. To feel the result, test three values, \(x = 0\), \(x = 7\), and \(x = 20\). How many of those three are solutions? Type the single count.
Show a hint
- Distribute the 3 on the left. \(3(2x + 4)\) becomes \(6x + 12\), which is exactly the right side.
- Since both sides are identical, every value of \(x\) is a solution, so all three test values work. Count them.
Show the full solution
Distribute the left side, \(3(2x + 4) = 6x + 12\), which is precisely the right side. Gathering the variable by subtracting \(6x\) leaves \(12 = 12\), a statement that is always true. That means every value of \(x\) is a solution, so the equation has infinitely many. In particular all three test values \(x = 0\), \(x = 7\), and \(x = 20\) work, which is \(\boxed{3}\) of the three. The equation is really just two ways of writing the same expression, the equivalent expressions you met in 6.1.
Problem
A shop compares two presses over \(t\) hours. The old press prints \(3(2t + 5)\) pages and the new one prints \(4t + 27\), and at one hour count they match. Distribute the left side, gather the variable, and find it. Solve \(3(2t + 5) = 4t + 27\) and type the value of \(t\).
Show a hint
- Distribute the 3 first. \(3(2t + 5)\) becomes \(6t + 15\), so the equation is \(6t + 15 = 4t + 27\).
- Subtract \(4t\) from both sides to get \(2t + 15 = 27\), then subtract 15 and divide by 2.
Show the full solution
Distribute the left side, \(3(2t + 5) = 6t + 15\), so the equation is \(6t + 15 = 4t + 27\). Gather the variable by subtracting \(4t\) from both sides, leaving \(2t + 15 = 27\). Subtract 15 to get \(2t = 12\), then divide by 2 to get \(t = 6\). Check, the old press gives \(3(2 \cdot 6 + 5) = 3(17) = 51\) and the new gives \(4(6) + 27 = 51\), so \(\boxed{6}\).
Practice these ideas
Practice
Solve \(9x = 5x + 12\) and type the value of \(x\).
Show the solution
Subtract \(5x\) from both sides, \(9x - 5x = 4x\), leaving \(4x = 12\). Divide both sides by 4 to get \(x = 3\). Check, the left is \(9(3) = 27\) and the right is \(5(3) + 12 = 27\), so \(\boxed{3}\).
Practice
Solve \(7y + 4 = 2y + 24\) and type the value of \(y\).
Show the solution
Subtract \(2y\) from both sides to get \(5y + 4 = 24\). Subtract 4 to get \(5y = 20\), then divide by 5 to get \(y = 4\). Check, the left is \(7(4) + 4 = 32\) and the right is \(2(4) + 24 = 32\), so \(\boxed{4}\).
Practice
First combine like terms on the left, then gather the variable. Solve \(7t + 2 - 3t = t + 17\) and type the value of \(t\).
Show the solution
Combine the like terms on the left, \(7t - 3t = 4t\), so the equation is \(4t + 2 = t + 17\). Subtract \(t\) from both sides to get \(3t + 2 = 17\), subtract 2 to get \(3t = 15\), then divide by 3 to get \(t = 5\). Check, the left is \(7(5) + 2 - 3(5) = 22\) and the right is \(5 + 17 = 22\), so \(\boxed{5}\).
Practice
Distribute the left side first, then gather the variable. Solve \(6(w - 2) = 3w + 6\) and type the value of \(w\).
Show the solution
Distribute the left side, \(6(w - 2) = 6w - 12\), so the equation is \(6w - 12 = 3w + 6\). Subtract \(3w\) from both sides to get \(3w - 12 = 6\). Add 12 to get \(3w = 18\), then divide by 3 to get \(w = 6\). Check, the left is \(6(6 - 2) = 24\) and the right is \(3(6) + 6 = 24\), so \(\boxed{6}\).
Practice
Solve \(7x + 1 = 2x + 9\) and type \(x\) as a fraction in lowest terms.
Show the solution
Subtract \(2x\) from both sides to get \(5x + 1 = 9\). Subtract 1 to get \(5x = 8\), then divide by 5 to get \(x = \tfrac{8}{5}\). Since 8 and 5 share no common factor, it is in lowest terms. Check, the left is \(7 \cdot \tfrac{8}{5} + 1 = \tfrac{56}{5} + \tfrac{5}{5} = \tfrac{61}{5}\) and the right is \(2 \cdot \tfrac{8}{5} + 9 = \tfrac{16}{5} + \tfrac{45}{5} = \tfrac{61}{5}\), so \(\boxed{8/5}\).
Practice
Clear the denominator first by multiplying every term by \(2\). Solve \(\dfrac{y}{2} + 5 = y - 1\) and type the value of \(y\).
Show the solution
Multiply every term on both sides by 2 to clear the fraction. The left \(\tfrac{y}{2} + 5\) becomes \(y + 10\), and the right \(y - 1\) becomes \(2y - 2\), so the equation is \(y + 10 = 2y - 2\). Subtract \(y\) from both sides to get \(10 = y - 2\), then add 2 to get \(y = 12\). Check, the left is \(\tfrac{12}{2} + 5 = 6 + 5 = 11\) and the right is \(12 - 1 = 11\), so \(\boxed{12}\).
Practice
Gather the variable in \(3x - 4 = 3x + 5\) by subtracting \(3x\) from both sides, then read what remains. How many solutions does it have? Type the single number.
Show the solution
Subtract \(3x\) from both sides. The variable cancels completely and leaves \(-4 = 5\), a false statement. The two sides always differ by 9, so no value of \(x\) can ever make them equal. The equation has no solution, a count of \(\boxed{0}\).
Practice
Distribute the left side of \(4(x + 2) = 4x + 8\) and compare it with the right. Test \(x = 1\), \(x = 9\), and \(x = 50\). How many of the three are solutions? Type the count.
Show the solution
Distribute the left side, \(4(x + 2) = 4x + 8\), which matches the right side exactly. Gathering the variable leaves \(8 = 8\), always true, so every value of \(x\) is a solution and there are infinitely many. All three test values \(x = 1\), \(x = 9\), and \(x = 50\) work, which is \(\boxed{3}\) of the three.
Practice
Two gyms charge by the month. Over \(m\) months gym A costs \(7m + 18\) and gym B costs \(9m\). Find the month count where they cost the same. Solve \(7m + 18 = 9m\) and type the value of \(m\).
Show the solution
Subtract \(7m\) from both sides to gather the variable on the right, leaving \(18 = 2m\). Divide both sides by 2 to get \(m = 9\). Check, gym A costs \(7(9) + 18 = 81\) and gym B costs \(9(9) = 81\), so they match at \(\boxed{9}\) months.