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Algebra · real student question

Solve for x: 0 = 2145 - 30*11*4 + 6*11*4 + 5*11*4 + x*11*10.

Question

Solve for xx:

0=2145−30⋅11⋅4+6⋅11⋅4+5⋅11⋅4+x⋅11⋅100=2145-30\cdot 11\cdot 4+6\cdot 11\cdot 4+5\cdot 11\cdot 4+x\cdot 11\cdot 10

Step-by-step solution

  1. Evaluate every product before combining anything. Multiplication outranks addition, so each triple product is a single constant (or a single coefficient of xx):

    30⋅11⋅4=1320,6⋅11⋅4=264,5⋅11⋅4=220,11⋅10=11030\cdot 11\cdot 4=1320,\quad 6\cdot 11\cdot 4=264,\quad 5\cdot 11\cdot 4=220,\quad 11\cdot 10=110

    Only the last one touches xx, so the equation has exactly one variable term with coefficient 110110.

  2. Rewrite with the numbers in place, keeping each sign attached.

    0=2145−1320+264+220+110x0=2145-1320+264+220+110x

    The minus belongs to the 13201320 term only; the two following products are added.

  3. Combine the constants left to right.

    2145−1320=825,825+264=1089,1089+220=13092145-1320=825,\qquad 825+264=1089,\qquad 1089+220=1309

    so the equation reduces to

    0=1309+110x0=1309+110x

  4. Isolate xx. Subtract 13091309 and divide by 110110:

    110x=−1309 ⟹ x=−1309110110x=-1309\ \Longrightarrow\ x=-\frac{1309}{110}

    The fraction simplifies to a terminating decimal because 1309=11⋅1191309=11\cdot 119 and 110=11⋅10110=11\cdot 10, so

    x=−11910=−11.9x=-\frac{119}{10}=-11.9

  5. Check by substitution. With x=−11.9x=-11.9: 110×(−11.9)=−1309110\times(-11.9)=-1309, and 1309+(−1309)=01309+(-1309)=0 ✓. The exact form −11910-\tfrac{119}{10} and the decimal −11.9-11.9 are the same number, so either is acceptable as the final answer.

Answer

x=−1309110=−11910=−11.9x=-\frac{1309}{110}=-\frac{119}{10}=-11.9

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