Find x in each equation: (25 + x) : 3 = 3^14 : 3^12; (x - 140) : 7 = 3^3 - 2^3 times 3; 54 : (x - 7) = 5^2 - 4^2; 12 times (x - 3) : 3 = 4^3 + 2^3; 45 : (3x - 4) = 3^2; (3x - 15) : 3 = 3^2; and (3x - 6) times 3 = 3^4.

Step-by-step solution

1. 1

   Always simplify the right-hand side to a single number first. Every right side is a power expression, and reducing it turns each equation into an ordinary one-or-two-step problem:

   $3^{14}:3^{12}=3^{2}=9,\quad 3^{3}-2^{3}\cdot 3=27-24=3,\quad 5^{2}-4^{2}=25-16=9,$
   $4^{3}+2^{3}=64+8=72,\quad 3^{2}=9,\quad 3^{4}=81.$

   Note in (2) that the power binds tighter than the multiplication: it is $27-(8\cdot 3)$, not $(27-8)\cdot 3$.

2. 2

   Equations 1 and 2: undo a division by a constant. Multiply both sides by the divisor, then subtract or add:

   $(25+x):3=9\Rightarrow 25+x=27\Rightarrow x=2,$
   $(x-140):7=3\Rightarrow x-140=21\Rightarrow x=161.$

3. 3

   Equation 3: the unknown sits in the divisor. When $x$ is underneath, swap the roles of divisor and quotient:

   $54:(x-7)=9\Rightarrow x-7=\frac{54}{9}=6\Rightarrow x=13.$

   The same idea applies to (5): $45:(3x-4)=9\Rightarrow 3x-4=5\Rightarrow x=3$.

4. 4

   Equation 4: two operations to unwind, in reverse order. The left side multiplies by $12$ then divides by $3$, so undo the division first:

   $12(x-3):3=72\Rightarrow 12(x-3)=216\Rightarrow x-3=18\Rightarrow x=21.$

5. 5

   Equations 6 and 7: unwind the outer operation, then the bracket.

   $(3x-15):3=9\Rightarrow 3x-15=27\Rightarrow 3x=42\Rightarrow x=14,$
   $(3x-6)\cdot 3=81\Rightarrow 3x-6=27\Rightarrow 3x=33\Rightarrow x=11.$

6. 6

   Collect and check the seven answers.

   $x=2,\ 161,\ 13,\ 21,\ 3,\ 14,\ 11.$

   Substituting back: $(25+2):3=9$ ✓, $(161-140):7=3$ ✓, $54:(13-7)=9$ ✓, $12(18):3=72$ ✓, $45:5=9$ ✓, $(42-15):3=9$ ✓, $(33-6)\cdot 3=81$ ✓.