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Ludmilka [50]
2 years ago
14

Solve the System of Equations -4x + 9y = 14 12x - 10y = - 8

Mathematics
2 answers:
12345 [234]2 years ago
6 0

Answer:

(1, 2)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

-4x + 9y = 14

12x - 10y = -8

<u>Step 2: Rewrite Systems</u>

-4x + 9y = 14

  1. Multiply everything by 3:                    -12x + 27y = 42

<u>Step 3: Redefine Systems</u>

-12x + 27y = 42

12x - 10y = -8

<u>Step 4: Solve for </u><em><u>y</u></em>

<em>Elimination</em>

  1. Combine 2 equations:                    17y = 34
  2. Divide 26 on both sides:                y = 2

<u>Step 5: Solve for </u><em><u>x</u></em>

  1. Define equation:                    12x - 10y = -8
  2. Substitute in <em>y</em>:                       12x - 10(2) = -8
  3. Multiply:                                  12x - 20 = -8
  4. Isolate <em>x</em> term:                        12x = 12
  5. Isolate <em>x</em>:                                 x = 1
WARRIOR [948]2 years ago
5 0

Step-by-step explanation:

HERE,

two equation are,

●-4x+9y=14••••••••••••(equation I)

●12x-10y=-8•••••••••••(equation II)

First multiplying 3 in equation I

we get,

\bold{3×(-4x+9y=14)   }

=\bold{ -12x+27y=42  }••(equation III)

Then,

we combine the equationii and equation III.

we get that,

\bold{12x-10y-12x+27y=-8+42   }

\bold{\cancel{12x}-10y\cancel{-12x}+27y=-8+42  }

\rightsquigarrow \bold{17y=34  }

\rightsquigarrow \bold{ y=\dfrac{34}{17}  }

\rightsquigarrow \boxed{  y=2 }

Then,

put the value of y in equation II.

WE get,

\rightsquigarrow \bold{12x-10×2=-8   }

\rightsquigarrow \bold{ 12x-20=-8  }

\rightsquigarrow \bold{ 12x=-8+20  }

\rightsquigarrow \bold{ 12x=12  }

\rightsquigarrow \bold{ x=\dfrac{12}{12}  }

\rightsquigarrow \boxed{ x=1  }

So,

<em>solution</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>two</em><em> </em><em>equation</em><em> </em><em>(</em><em>-4x</em><em>+</em><em>9y</em><em>)</em><em> </em><em>and</em><em> </em><em>(</em><em>12x-10y</em><em>=</em><em>-8</em><em>)</em> <em>is</em><em> </em><em><u>(</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>)</u></em>

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