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Activity Discussion Math Maths Reply To: Maths

• ### brajesh

Member
July 10, 2024 at 12:50 pm
0
1. Distribute the 3 on the left side of the equation:

<math xmlns=”http://www.w3.org/1998/Math/MathML&#8221; display=”block”><semantics><mrow><mn>3</mn><mo stretchy=”false”>(</mo><mi>y</mi><mo>−</mo><mn>4</mn><mo stretchy=”false”>)</mo><mo>=</mo><mn>15</mn><mtext>  </mtext><mo>⟹</mo><mtext>  </mtext><mn>3</mn><mi>y</mi><mo>−</mo><mn>12</mn><mo>=</mo><mn>15</mn></mrow><annotation encoding=”application/x-tex”>3(y – 4) = 15 \implies 3y – 12 = 15</annotation></semantics>[/itex]3(y−4)=15⟹3y−12=15

2. Add 12 to both sides of the equation to isolate the term with <math xmlns=”http://www.w3.org/1998/Math/MathML”><semantics><mrow><mi>y</mi></mrow><annotation encoding=”application/x-tex”>y</annotation></semantics>[/itex]y:

<math xmlns=”http://www.w3.org/1998/Math/MathML&#8221; display=”block”><semantics><mrow><mn>3</mn><mi>y</mi><mo>−</mo><mn>12</mn><mo>+</mo><mn>12</mn><mo>=</mo><mn>15</mn><mo>+</mo><mn>12</mn><mtext>  </mtext><mo>⟹</mo><mtext>  </mtext><mn>3</mn><mi>y</mi><mo>=</mo><mn>27</mn></mrow><annotation encoding=”application/x-tex”>3y – 12 + 12 = 15 + 12 \implies 3y = 27</annotation></semantics>[/itex]3y−12+12=15+12⟹3y=27

3. Divide both sides by 3 to solve for <math xmlns=”http://www.w3.org/1998/Math/MathML”><semantics><mrow><mi>y</mi></mrow><annotation encoding=”application/x-tex”>y</annotation></semantics>[/itex]y:

<math xmlns=”http://www.w3.org/1998/Math/MathML&#8221; display=”block”><semantics><mrow><mfrac><mrow><mn>3</mn><mi>y</mi></mrow><mn>3</mn></mfrac><mo>=</mo><mfrac><mn>27</mn><mn>3</mn></mfrac><mtext>  </mtext><mo>⟹</mo><mtext>  </mtext><mi>y</mi><mo>=</mo><mn>9</mn></mrow><annotation encoding=”application/x-tex”>\frac{3y}{3} = \frac{27}{3} \implies y = 9</annotation></semantics>[/itex]33y​=327​⟹y=9

So, the solution is <math xmlns=”http://www.w3.org/1998/Math/MathML”><semantics><mrow><mi>y</mi><mo>=</mo><mn>9</mn></mrow><annotation encoding=”application/x-tex”>y = 9</annotation></semantics>[/itex]y=9.

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