Answer :
Sure, let's solve these step-by-step and find four solutions for each given equation.
### (i) [tex]\( 2x + y = 7 \)[/tex]
To generate solutions for this linear equation, we can choose different values of [tex]\( x \)[/tex] and then solve for [tex]\( y \)[/tex].
Let's choose [tex]\( x \)[/tex] values from 1 to 4:
1. When [tex]\( x = 1 \)[/tex]:
[tex]\[ 2(1) + y = 7 \implies 2 + y = 7 \implies y = 5 \][/tex]
Solution: [tex]\( (1, 5) \)[/tex]
2. When [tex]\( x = 2 \)[/tex]:
[tex]\[ 2(2) + y = 7 \implies 4 + y = 7 \implies y = 3 \][/tex]
Solution: [tex]\( (2, 3) \)[/tex]
3. When [tex]\( x = 3 \)[/tex]:
[tex]\[ 2(3) + y = 7 \implies 6 + y = 7 \implies y = 1 \][/tex]
Solution: [tex]\( (3, 1) \)[/tex]
4. When [tex]\( x = 4 \)[/tex]:
[tex]\[ 2(4) + y = 7 \implies 8 + y = 7 \implies y = -1 \][/tex]
Solution: [tex]\( (4, -1) \)[/tex]
So, the four solutions for [tex]\( 2x + y = 7 \)[/tex] are:
[tex]\[ (1, 5), (2, 3), (3, 1), (4, -1) \][/tex]
### (ii) [tex]\( \pi x + y = 9 \)[/tex]
For this equation, we use the value of [tex]\( \pi \approx 3.14159 \)[/tex].
Let's choose [tex]\( x \)[/tex] values from 1 to 4:
1. When [tex]\( x = 1 \)[/tex]:
[tex]\[ \pi(1) + y = 9 \implies 3.14159 + y = 9 \implies y \approx 9 - 3.14159 \implies y \approx 5.85841 \][/tex]
Solution: [tex]\( (1, 5.85841) \)[/tex]
2. When [tex]\( x = 2 \)[/tex]:
[tex]\[ \pi(2) + y = 9 \implies 6.28318 + y = 9 \implies y \approx 9 - 6.28318 \implies y \approx 2.71681 \][/tex]
Solution: [tex]\( (2, 2.71681) \)[/tex]
3. When [tex]\( x = 3 \)[/tex]:
[tex]\[ \pi(3) + y = 9 \implies 9.42477 + y = 9 \implies y \approx 9 - 9.42477 \implies y \approx -0.42478 \][/tex]
Solution: [tex]\( (3, -0.42478) \)[/tex]
4. When [tex]\( x = 4 \)[/tex]:
[tex]\[ \pi(4) + y = 9 \implies 12.56636 + y = 9 \implies y \approx 9 - 12.56636 \implies y \approx -3.56637 \][/tex]
Solution: [tex]\( (4, -3.56637) \)[/tex]
So, the four solutions for [tex]\( \pi x + y = 9 \)[/tex] are:
[tex]\[ (1, 5.85841), (2, 2.71681), (3, -0.42478), (4, -3.56637) \][/tex]
### (iii) [tex]\( x = 4y \)[/tex]
For this equation, we can choose different values of [tex]\( y \)[/tex] and solve for [tex]\( x \)[/tex].
Let's choose [tex]\( y \)[/tex] values from 1 to 4:
1. When [tex]\( y = 1 \)[/tex]:
[tex]\[ x = 4(1) \implies x = 4 \][/tex]
Solution: [tex]\( (4, 1) \)[/tex]
2. When [tex]\( y = 2 \)[/tex]:
[tex]\[ x = 4(2) \implies x = 8 \][/tex]
Solution: [tex]\( (8, 2) \)[/tex]
3. When [tex]\( y = 3 \)[/tex]:
[tex]\[ x = 4(3) \implies x = 12 \][/tex]
Solution: [tex]\( (12, 3) \)[/tex]
4. When [tex]\( y = 4 \)[/tex]:
[tex]\[ x = 4(4) \implies x = 16 \][/tex]
Solution: [tex]\( (16, 4) \)[/tex]
So, the four solutions for [tex]\( x = 4y \)[/tex] are:
[tex]\[ (4, 1), (8, 2), (12, 3), (16, 4) \][/tex]
### Combined Solutions
By following the process for each equation, we get the following points for each equation:
[tex]\[ \begin{align*} (i) &\quad (1, 5), (2, 3), (3, 1), (4, -1) \\ (ii) &\quad (1, 5.85841), (2, 2.71681), (3, -0.42478), (4, -3.56637) \\ (iii) &\quad (4, 1), (8, 2), (12, 3), (16, 4) \end{align*} \][/tex]
These points represent the four solutions for each given equation.
### (i) [tex]\( 2x + y = 7 \)[/tex]
To generate solutions for this linear equation, we can choose different values of [tex]\( x \)[/tex] and then solve for [tex]\( y \)[/tex].
Let's choose [tex]\( x \)[/tex] values from 1 to 4:
1. When [tex]\( x = 1 \)[/tex]:
[tex]\[ 2(1) + y = 7 \implies 2 + y = 7 \implies y = 5 \][/tex]
Solution: [tex]\( (1, 5) \)[/tex]
2. When [tex]\( x = 2 \)[/tex]:
[tex]\[ 2(2) + y = 7 \implies 4 + y = 7 \implies y = 3 \][/tex]
Solution: [tex]\( (2, 3) \)[/tex]
3. When [tex]\( x = 3 \)[/tex]:
[tex]\[ 2(3) + y = 7 \implies 6 + y = 7 \implies y = 1 \][/tex]
Solution: [tex]\( (3, 1) \)[/tex]
4. When [tex]\( x = 4 \)[/tex]:
[tex]\[ 2(4) + y = 7 \implies 8 + y = 7 \implies y = -1 \][/tex]
Solution: [tex]\( (4, -1) \)[/tex]
So, the four solutions for [tex]\( 2x + y = 7 \)[/tex] are:
[tex]\[ (1, 5), (2, 3), (3, 1), (4, -1) \][/tex]
### (ii) [tex]\( \pi x + y = 9 \)[/tex]
For this equation, we use the value of [tex]\( \pi \approx 3.14159 \)[/tex].
Let's choose [tex]\( x \)[/tex] values from 1 to 4:
1. When [tex]\( x = 1 \)[/tex]:
[tex]\[ \pi(1) + y = 9 \implies 3.14159 + y = 9 \implies y \approx 9 - 3.14159 \implies y \approx 5.85841 \][/tex]
Solution: [tex]\( (1, 5.85841) \)[/tex]
2. When [tex]\( x = 2 \)[/tex]:
[tex]\[ \pi(2) + y = 9 \implies 6.28318 + y = 9 \implies y \approx 9 - 6.28318 \implies y \approx 2.71681 \][/tex]
Solution: [tex]\( (2, 2.71681) \)[/tex]
3. When [tex]\( x = 3 \)[/tex]:
[tex]\[ \pi(3) + y = 9 \implies 9.42477 + y = 9 \implies y \approx 9 - 9.42477 \implies y \approx -0.42478 \][/tex]
Solution: [tex]\( (3, -0.42478) \)[/tex]
4. When [tex]\( x = 4 \)[/tex]:
[tex]\[ \pi(4) + y = 9 \implies 12.56636 + y = 9 \implies y \approx 9 - 12.56636 \implies y \approx -3.56637 \][/tex]
Solution: [tex]\( (4, -3.56637) \)[/tex]
So, the four solutions for [tex]\( \pi x + y = 9 \)[/tex] are:
[tex]\[ (1, 5.85841), (2, 2.71681), (3, -0.42478), (4, -3.56637) \][/tex]
### (iii) [tex]\( x = 4y \)[/tex]
For this equation, we can choose different values of [tex]\( y \)[/tex] and solve for [tex]\( x \)[/tex].
Let's choose [tex]\( y \)[/tex] values from 1 to 4:
1. When [tex]\( y = 1 \)[/tex]:
[tex]\[ x = 4(1) \implies x = 4 \][/tex]
Solution: [tex]\( (4, 1) \)[/tex]
2. When [tex]\( y = 2 \)[/tex]:
[tex]\[ x = 4(2) \implies x = 8 \][/tex]
Solution: [tex]\( (8, 2) \)[/tex]
3. When [tex]\( y = 3 \)[/tex]:
[tex]\[ x = 4(3) \implies x = 12 \][/tex]
Solution: [tex]\( (12, 3) \)[/tex]
4. When [tex]\( y = 4 \)[/tex]:
[tex]\[ x = 4(4) \implies x = 16 \][/tex]
Solution: [tex]\( (16, 4) \)[/tex]
So, the four solutions for [tex]\( x = 4y \)[/tex] are:
[tex]\[ (4, 1), (8, 2), (12, 3), (16, 4) \][/tex]
### Combined Solutions
By following the process for each equation, we get the following points for each equation:
[tex]\[ \begin{align*} (i) &\quad (1, 5), (2, 3), (3, 1), (4, -1) \\ (ii) &\quad (1, 5.85841), (2, 2.71681), (3, -0.42478), (4, -3.56637) \\ (iii) &\quad (4, 1), (8, 2), (12, 3), (16, 4) \end{align*} \][/tex]
These points represent the four solutions for each given equation.
