Activity › Discussion › Math › Maths › Reply To: Maths
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Distribute the 3 on the left side of the equation:
<math xmlns=”http://www.w3.org/1998/Math/MathML” 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></math>3(y−4)=15⟹3y−12=15
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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></math>y:
<math xmlns=”http://www.w3.org/1998/Math/MathML” 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></math>3y−12+12=15+12⟹3y=27
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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></math>y:
<math xmlns=”http://www.w3.org/1998/Math/MathML” 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></math>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></math>y=9.
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