Answer :
To solve the equation [tex]\( b \cdot 4b + a - 3 = b + 5a \)[/tex], we need to verify which option for [tex]\( b \)[/tex] makes the equation true. Let's analyze each option step-by-step:
### Option 1: [tex]\( b = \frac{4a+3}{3} \)[/tex]
1. Substitute [tex]\( b = \frac{4a+3}{3} \)[/tex] into the equation:
[tex]\[ \left( \frac{4a+3}{3} \right) \cdot 4 \left( \frac{4a+3}{3} \right) + a - 3 = \frac{4a+3}{3} + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{4a+3}{3} \right) \cdot \left( \frac{16a^2 + 24a + 9}{9} \right) + a - 3 \][/tex]
Simplify further:
[tex]\[ \frac{16a^2 + 24a + 9}{9} + a - 3 \][/tex]
To get a common denominator:
[tex]\[ \frac{16a^2 + 24a + 9 + 9a - 27}{9} = \frac{16a^2 + 33a - 18}{9} \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{4a+3}{3} + 5a = \frac{4a+3 + 15a}{3} = \frac{19a + 3}{3} \][/tex]
4. Compare both sides:
[tex]\[ \frac{16a^2 + 33a - 18}{9} \neq \frac{19a+3}{3} \][/tex]
So, option 1 is incorrect.
### Option 2: [tex]\( b = \frac{4a+3}{4} \)[/tex]
1. Substitute [tex]\( b = \frac{4a+3}{4} \)[/tex] into the equation:
[tex]\[ \left( \frac{4a+3}{4} \right) \cdot 4 \left( \frac{4a+3}{4} \right) + a - 3 = \frac{4a+3}{4} + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{4a+3}{4} \right) \cdot \left( \frac{16a^2 + 24a + 9}{16} \right) + a - 3 \][/tex]
Simplify further:
[tex]\[ \frac{16a^2 + 24a + 9}{16} + a - 3 = \frac{16a^2 + 24a + 9 + 16a - 48}{16} = \frac{16a^2 + 40a - 39}{16} \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{4a+3}{4} + 5a = \frac{4a+3 + 20a}{4} = \frac{24a + 3}{4} \][/tex]
4. Compare both sides:
[tex]\[ \frac{16a^2 + 40a - 39}{16} \neq \frac{24a+3}{4} \][/tex]
So, option 2 is incorrect.
### Option 3: [tex]\( b = \frac{4a}{3} + 3 \)[/tex]
1. Substitute [tex]\( b = \frac{4a}{3} + 3 \)[/tex] into the equation:
[tex]\[ \left( \frac{4a}{3} + 3 \right) \cdot 4 \left( \frac{4a}{3} + 3 \right) + a - 3 = \left( \frac{4a}{3} + 3 \right) + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{4a}{3} + 3 \right) \cdot \left( \frac{16a^2}{9} + 16a + 36 \right) + a - 3 \][/tex]
Simplify further:
[tex]\[ \frac{16a^2}{9} + 16a + 36 + a - 3 = \frac{16a^2}{9} + 17a + 33 = \frac{16a^2 + 153a + 297}{27} \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{4a}{3} + 3 + 5a = \frac{4a + 9a + 9}{3} = \frac{13a + 9}{3} \][/tex]
4. Compare both sides:
[tex]\[ \frac{16a^2 + 153a + 297}{27} \neq \frac{13a + 9}{3} \][/tex]
So, option 3 is incorrect.
### Option 4: [tex]\( b = \frac{7a}{3} \)[/tex]
1. Substitute [tex]\( b = \frac{7a}{3} \)[/tex] into the equation:
[tex]\[ \left( \frac{7a}{3} \right) \cdot 4 \left( \frac{7a}{3} \right) + a - 3 = \frac{7a}{3} + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{7a}{3} \right) \cdot \left( \frac{28a^2}{9} \right) + a - 3 = \frac{196a^2}{9} + a - 3 \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{7a}{3} + 5a = \frac{7a + 15a}{3} = \frac{22a}{3} \][/tex]
4. Compare both sides:
[tex]\[ \frac{196a^2}{9} + a - 3 \neq \frac{22a}{3} \][/tex]
So, option 4 is incorrect.
Since none of the options satisfies the initial equation, it's verified that there may have been an incomplete or incorrect setup in the options provided. Therefore, none of the given answers correctly solves the equation.
### Option 1: [tex]\( b = \frac{4a+3}{3} \)[/tex]
1. Substitute [tex]\( b = \frac{4a+3}{3} \)[/tex] into the equation:
[tex]\[ \left( \frac{4a+3}{3} \right) \cdot 4 \left( \frac{4a+3}{3} \right) + a - 3 = \frac{4a+3}{3} + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{4a+3}{3} \right) \cdot \left( \frac{16a^2 + 24a + 9}{9} \right) + a - 3 \][/tex]
Simplify further:
[tex]\[ \frac{16a^2 + 24a + 9}{9} + a - 3 \][/tex]
To get a common denominator:
[tex]\[ \frac{16a^2 + 24a + 9 + 9a - 27}{9} = \frac{16a^2 + 33a - 18}{9} \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{4a+3}{3} + 5a = \frac{4a+3 + 15a}{3} = \frac{19a + 3}{3} \][/tex]
4. Compare both sides:
[tex]\[ \frac{16a^2 + 33a - 18}{9} \neq \frac{19a+3}{3} \][/tex]
So, option 1 is incorrect.
### Option 2: [tex]\( b = \frac{4a+3}{4} \)[/tex]
1. Substitute [tex]\( b = \frac{4a+3}{4} \)[/tex] into the equation:
[tex]\[ \left( \frac{4a+3}{4} \right) \cdot 4 \left( \frac{4a+3}{4} \right) + a - 3 = \frac{4a+3}{4} + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{4a+3}{4} \right) \cdot \left( \frac{16a^2 + 24a + 9}{16} \right) + a - 3 \][/tex]
Simplify further:
[tex]\[ \frac{16a^2 + 24a + 9}{16} + a - 3 = \frac{16a^2 + 24a + 9 + 16a - 48}{16} = \frac{16a^2 + 40a - 39}{16} \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{4a+3}{4} + 5a = \frac{4a+3 + 20a}{4} = \frac{24a + 3}{4} \][/tex]
4. Compare both sides:
[tex]\[ \frac{16a^2 + 40a - 39}{16} \neq \frac{24a+3}{4} \][/tex]
So, option 2 is incorrect.
### Option 3: [tex]\( b = \frac{4a}{3} + 3 \)[/tex]
1. Substitute [tex]\( b = \frac{4a}{3} + 3 \)[/tex] into the equation:
[tex]\[ \left( \frac{4a}{3} + 3 \right) \cdot 4 \left( \frac{4a}{3} + 3 \right) + a - 3 = \left( \frac{4a}{3} + 3 \right) + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{4a}{3} + 3 \right) \cdot \left( \frac{16a^2}{9} + 16a + 36 \right) + a - 3 \][/tex]
Simplify further:
[tex]\[ \frac{16a^2}{9} + 16a + 36 + a - 3 = \frac{16a^2}{9} + 17a + 33 = \frac{16a^2 + 153a + 297}{27} \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{4a}{3} + 3 + 5a = \frac{4a + 9a + 9}{3} = \frac{13a + 9}{3} \][/tex]
4. Compare both sides:
[tex]\[ \frac{16a^2 + 153a + 297}{27} \neq \frac{13a + 9}{3} \][/tex]
So, option 3 is incorrect.
### Option 4: [tex]\( b = \frac{7a}{3} \)[/tex]
1. Substitute [tex]\( b = \frac{7a}{3} \)[/tex] into the equation:
[tex]\[ \left( \frac{7a}{3} \right) \cdot 4 \left( \frac{7a}{3} \right) + a - 3 = \frac{7a}{3} + 5a \][/tex]
2. Evaluate the left side:
[tex]\[ \left( \frac{7a}{3} \right) \cdot \left( \frac{28a^2}{9} \right) + a - 3 = \frac{196a^2}{9} + a - 3 \][/tex]
3. Evaluate the right side:
[tex]\[ \frac{7a}{3} + 5a = \frac{7a + 15a}{3} = \frac{22a}{3} \][/tex]
4. Compare both sides:
[tex]\[ \frac{196a^2}{9} + a - 3 \neq \frac{22a}{3} \][/tex]
So, option 4 is incorrect.
Since none of the options satisfies the initial equation, it's verified that there may have been an incomplete or incorrect setup in the options provided. Therefore, none of the given answers correctly solves the equation.
