Solving Linear Equations

The +8 sits next to x; the inverse of adding 8 is subtracting 8. Do it to both sides so the scale stays level:
$$x+8=15 \;\xrightarrow{\,-8\,}\; x=7$$
The +8 goes to zero, leaving x alone. Check in the original: 7 + 8 = 15.

Answer: 7

We are subtracting 13 from x, so undo it by adding 13 to both sides:
$$x-13=4 \;\xrightarrow{\,+13\,}\; x=17$$
The −13 goes to zero. Check: 17 − 13 = 4.

Answer: 17

Here x is multiplied by 7, so the inverse is dividing both sides by 7:
$$7x=63 \;\xrightarrow{\,\div 7\,}\; x=9$$
The coefficient goes to one because 7/7 = 1, and 1x = x. Check: 7(9) = 63.

Answer: 9

x is divided by 4, so undo it by multiplying both sides by 4:
$$\frac{x}{4}=9 \;\xrightarrow{\,\times 4\,}\; x=36$$
Dividing by 4 then multiplying by 4 takes the left side back to x (it goes to one times x). Check: 36/4 = 9. Notice that dividing made x larger than 9, not smaller — the original was already cut into quarters.

Answer: 36

This was built by multiplying x by 3, then adding 5 — so undo in the opposite order: shoes off first (subtract 5), then socks (divide by 3).
$$3x+5=23 \;\xrightarrow{\,-5\,}\; 3x=18 \;\xrightarrow{\,\div 3\,}\; x=6$$
The +5 is on the outside, so it comes off first; stripping it keeps the numbers whole. Check: 3(6) + 5 = 18 + 5 = 23.

Answer: 6

Undo the −8 first (add 8 to both sides), then undo the ×5 (divide by 5):
$$5x-8=32 \;\xrightarrow{\,+8\,}\; 5x=40 \;\xrightarrow{\,\div 5\,}\; x=8$$
Check: 5(8) − 8 = 40 − 8 = 32.

Answer: 8

Strip the +4 first (subtract 4), then undo the ÷3 (multiply by 3):
$$\frac{x}{3}+4=10 \;\xrightarrow{\,-4\,}\; \frac{x}{3}=6 \;\xrightarrow{\,\times 3\,}\; x=18$$
Check: 18/3 + 4 = 6 + 4 = 10.

Answer: 18

Keep the operation sign separate from the number's sign: the term is −3x. Subtract 9 from both sides, then divide by −3.
$$9-3x=21 \;\xrightarrow{\,-9\,}\; -3x=12 \;\xrightarrow{\,\div(-3)\,}\; x=-4$$
Dividing a positive by a negative gives a negative: 12 / (−3) = −4. Check (this is where a sign slip would show): 9 − 3(−4) = 9 + 12 = 21. A negative solution is a real, valid answer.

Answer: -4

These are like terms — both are the same kind of box (x). Add the coefficients: 6 + 5 = 11, so 6x + 5x = 11x.

Answer: 11*x

Group like with like. The x-terms: 8x − 2x = 6x. The constants: −3 + 2 = −1. They are different kinds of thing (boxes vs. loose units), so you cannot merge them further:
$$8x-3+2-2x = 6x-1$$

Answer: 6*x-1

The outside 4 must hand a flyer to everyone inside — nobody skipped. 4·x = 4x and 4·6 = 24:
$$4(x+6)=4x+24$$
Area-model picture: a rectangle 4 tall and (x + 6) wide splits into 4x and 24.

Answer: 4*x+24

The break is in the distribution. The −2 must greet every term inside, including the −3: −2 × −3 = +6 (a negative times a negative is positive), not −6. So the correct second line is 6 − 2x + 6, which combines to −2x + 12 — not −2x.

Confirm with x = 1: the original is 6 − 2(1 − 3) = 6 − 2(−2) = 6 + 4 = 10. The student's −2x gives −2(1) = −2 (wrong). The corrected −2x + 12 gives −2(1) + 12 = 10.

Answer: The error is in distributing the negative: −2 × −3 should be +6, not −6. Correct: 6 − 2(x − 3) = 6 − 2x + 6 = −2x + 12.

Both pans have boxes and coins. Move the smaller variable term (4x) to avoid negatives — subtract 4x from both sides (the 4x on the right goes to zero):
$$7x+2=4x+17 \;\xrightarrow{\,-4x\,}\; 3x+2=17 \;\xrightarrow{\,-2\,}\; 3x=15 \;\xrightarrow{\,\div 3\,}\; x=5$$
Check: 7(5) + 2 = 37 and 4(5) + 17 = 37.

Answer: 5

Subtract the smaller variable term, 2x, from both sides, then gather constants:
$$5x-3=2x+9 \;\xrightarrow{\,-2x\,}\; 3x-3=9 \;\xrightarrow{\,+3\,}\; 3x=12 \;\xrightarrow{\,\div 3\,}\; x=4$$
Check: 5(4) − 3 = 17 and 2(4) + 9 = 17.

Answer: 4

Tidy each side first: the left side has a factor wrapped around (x + 4), so distribute the 3 before any terms can move across.
$$3(x+4)=x+20 \;\xrightarrow{\,\text{distribute}\,}\; 3x+12=x+20 \;\xrightarrow{\,-x\,}\; 2x+12=20 \;\xrightarrow{\,-12\,}\; 2x=8 \;\xrightarrow{\,\div 2\,}\; x=4$$
You distribute first because the 3 multiplies the whole group (x + 4); until it is handed out, the 3x is still locked inside the parentheses and you cannot legally combine it with the x on the right. Check: 3(4 + 4) = 24 and 4 + 20 = 24.

Answer: 4

Distribute the left side: 3(2x + 4) = 6x + 12. Now the equation reads 6x + 12 = 6x + 12 — the two sides are the same expression. Gather the variable terms (subtract 6x from both sides):
$$6x+12=6x+12 \;\xrightarrow{\,-6x\,}\; 12=12$$
The variable vanished and left 12 = 12, a true statement. So every number works → identity, all real numbers (infinitely many solutions). Sanity check with two values: at x = 0 both sides are 12; at x = 5 both sides are 42. They match for any x. (Don't write 'x = 12 = 12' — once x is gone there is no single value to report.)

Answer: Identity — all real numbers (infinitely many solutions).

The reciprocal of 1/3 is 3/1 = 3. Multiply both sides by 3 (this makes the coefficient go to one, since (1/3)·3 = 1):
$$\tfrac13 x=6 \;\xrightarrow{\,\times 3\,}\; x=18$$
Check: (1/3)(18) = 6.

Answer: 18

The denominators are 2 and 3; the LCM is 6, the smallest number both divide. Multiply every term on both sides by 6 (the balance stays level only if the whole equation is multiplied):
$$\frac{x}{2}+\frac{x}{3}=10 \;\xrightarrow{\,\times 6\,}\; 3x+2x=60 \;\Rightarrow\; 5x=60 \;\Rightarrow\; x=12$$
Check: 12/2 + 12/3 = 6 + 4 = 10.

Answer: 12

First strip the +2 (subtract 2 from both sides), then undo the (3/4) coefficient by multiplying by its reciprocal 4/3:
$$\tfrac34 x+2=11 \;\xrightarrow{\,-2\,}\; \tfrac34 x=9 \;\xrightarrow{\,\times \tfrac43\,}\; x=9\cdot\tfrac43=12$$
Check: (3/4)(12) + 2 = 9 + 2 = 11.

Answer: 12

Multiply both sides by the reciprocal of 2/3, which is 3/2:
$$\tfrac23 x=7 \;\xrightarrow{\,\times \tfrac32\,}\; x=7\cdot\tfrac32=\frac{21}{2}$$
An exact fraction is the answer — 21/2 (that is 10½); don't round to 10 or 11. Check: (2/3)·(21/2) = 42/6 = 7.

Answer: 21/2

The variable term is −2x. Subtract 6 from both sides, then divide by −2:
$$6-2x=16 \;\xrightarrow{\,-6\,}\; -2x=10 \;\xrightarrow{\,\div(-2)\,}\; x=-5$$
Check: 6 − 2(−5) = 6 + 10 = 16. A negative answer is perfectly valid.

Answer: -5

Distribute −3 to every term inside: −3·x = −3x and −3·(−2) = +6 (negative times negative is positive). Then combine constants:
$$4-3(x-2) = 4-3x+6 = -3x+10$$
A common wrong answer is 4 − 3x − 6 = −3x − 2, from giving −3 × −2 the wrong sign. Check at x = 1: original 4 − 3(1 − 2) = 4 − 3(−1) = 7; rewrite −3(1) + 10 = 7.

Answer: -3*x+10

Distribute the 2 (positive 2 over a subtraction keeps the −8): 2(x − 4) = 2x − 8. Then gather:
$$2(x-4)=x+1 \;\xrightarrow{\,\text{distribute}\,}\; 2x-8=x+1 \;\xrightarrow{\,-x\,}\; x-8=1 \;\xrightarrow{\,+8\,}\; x=9$$
Check: 2(9 − 4) = 2(5) = 10 and 9 + 1 = 10.

Answer: 9

Denominators 4 and 6 have LCM 12. Multiply every term by 12:
$$\frac{x}{4}+\frac{x}{6}=5 \;\xrightarrow{\,\times 12\,}\; 3x+2x=60 \;\Rightarrow\; 5x=60 \;\Rightarrow\; x=12$$
Check: 12/4 + 12/6 = 3 + 2 = 5.

Answer: 12

Distribute the right side: 2(3x + 5) = 6x + 10. The equation becomes 6x + 4 = 6x + 10. Gather the variable terms (subtract 6x from both sides):
$$6x+4=6x+10 \;\xrightarrow{\,-6x\,}\; 4=10$$
The variable vanished and left 4 = 10, which is false. No number can make a false statement true → contradiction, no solution. Sanity check: at x = 0 the sides are 4 and 10; the right side is always 6 more, so they can never be equal. (Don't write 'x = 4 = 10' — there is no x left to report.)

Answer: Contradiction — no solution.