Math, asked by jyotirmaykalitajishn, 7 months ago

write any four solutions for the equations. a. x + 2y = -5. b. 3x + y = 7. c. 3x + 5y = 15​

Answers

Answered by Anonymous
3

Answer:-

a)x + 2y =  - 5 \\1st \:  \: solution \:   \\ x = 1 \\  \\ 1 + 2y =  - 5 \\  \\ 2y =  - 5 - 1 \\  \\ 2y =  - 6 \\  \\ y =  \frac{ - 6}{2}  \\  \\ y =  - 3 \\ 2nd \:  \:  \: solution \:  \\ x = 2 \\  \\ 2 + 2y =  - 5 \\  \\ 2y =  - 7 \\  \\ y =  \frac{ - 7}{2}  \\  3rd \:  \: solution\\ x = 3 \\  \\ 3 + 2y =  - 5 \\  \\ 2y = 8 \\  \\ y =  - 4 \\  \\ 4th \:  \: solution \\  \\ y = 1 \\  \\ x + 2 =  - 5 \\  \\ x =  - 7

b)3x + y = 7 \\  first \:  \: solution\\ y = 1 \\  \\ 3x + 1 = 7 \\  \\ 3x = 6 \\  \\ x = 2 \\ 2 \: nd \:  \: solution \:  \\  \\ y = 2 \\ 3x + 2 = 7 \\  \\ 3x = 5 \\  \\ x =  \frac{5}{3}  \\  \\ 3rd \:  \: solution \:  \\  \\ y = 3 \\  \\ 3x + 3 = 7 \\  \\ 3x = 4 \\  \\ x =  \frac{4}{3}  \\  \\ 4th \:  \: solution \\  \\ x = 1 \\  \\ 3 + y = 7 \\  \\ y = 4

c)3x + 5y = 15 \\  \\ 1 \: st \:  \: solution \\  \\ x = 1 \\  \\ 3 + 5y = 15 \\  \\ 5y = 12 \\  \\ y = 12 \div 5 \\  \\ 2nd \:  \: solution \\  \\ x = 2 \\  \\ 6 + 5y = 15 \\  \\ 5y = 9 \\  \\ y = 9 \div 5 \\  \\ 3rd \:  \: solution \\  \\ y = 1 \\  \\ 3x + 5 = 15 \\  \\ 3x = 10 \\  \\ x = 10 \div 3 \\  \\ 4th \:  \: solution \\  \\ y = 2 \\  \\ 3x + 10 = 15 \\  \\ 3x = 5 \\  \\ x = 5 \div 3

Answered by meeenvemn
0

Step-by-step explanation:

Answer:-

\begin{gathered}a)x + 2y = - 5 \\1st \: \: solution \: \\ x = 1 \\ \\ 1 + 2y = - 5 \\ \\ 2y = - 5 - 1 \\ \\ 2y = - 6 \\ \\ y = \frac{ - 6}{2} \\ \\ y = - 3 \\ 2nd \: \: \: solution \: \\ x = 2 \\ \\ 2 + 2y = - 5 \\ \\ 2y = - 7 \\ \\ y = \frac{ - 7}{2} \\ 3rd \: \: solution\\ x = 3 \\ \\ 3 + 2y = - 5 \\ \\ 2y = 8 \\ \\ y = - 4 \\ \\ 4th \: \: solution \\ \\ y = 1 \\ \\ x + 2 = - 5 \\ \\ x = - 7\end{gathered}

a)x+2y=−5

1stsolution

x=1

1+2y=−5

2y=−5−1

2y=−6

y=

2

−6

y=−3

2ndsolution

x=2

2+2y=−5

2y=−7

y=

2

−7

3rdsolution

x=3

3+2y=−5

2y=8

y=−4

4thsolution

y=1

x+2=−5

x=−7

\begin{gathered}b)3x + y = 7 \\ first \: \: solution\\ y = 1 \\ \\ 3x + 1 = 7 \\ \\ 3x = 6 \\ \\ x = 2 \\ 2 \: nd \: \: solution \: \\ \\ y = 2 \\ 3x + 2 = 7 \\ \\ 3x = 5 \\ \\ x = \frac{5}{3} \\ \\ 3rd \: \: solution \: \\ \\ y = 3 \\ \\ 3x + 3 = 7 \\ \\ 3x = 4 \\ \\ x = \frac{4}{3} \\ \\ 4th \: \: solution \\ \\ x = 1 \\ \\ 3 + y = 7 \\ \\ y = 4\end{gathered}

b)3x+y=7

firstsolution

y=1

3x+1=7

3x=6

x=2

2ndsolution

y=2

3x+2=7

3x=5

x=

3

5

3rdsolution

y=3

3x+3=7

3x=4

x=

3

4

4thsolution

x=1

3+y=7

y=4

\begin{gathered}c)3x + 5y = 15 \\ \\ 1 \: st \: \: solution \\ \\ x = 1 \\ \\ 3 + 5y = 15 \\ \\ 5y = 12 \\ \\ y = 12 \div 5 \\ \\ 2nd \: \: solution \\ \\ x = 2 \\ \\ 6 + 5y = 15 \\ \\ 5y = 9 \\ \\ y = 9 \div 5 \\ \\ 3rd \: \: solution \\ \\ y = 1 \\ \\ 3x + 5 = 15 \\ \\ 3x = 10 \\ \\ x = 10 \div 3 \\ \\ 4th \: \: solution \\ \\ y = 2 \\ \\ 3x + 10 = 15 \\ \\ 3x = 5 \\ \\ x = 5 \div 3\end{gathered}

c)3x+5y=15

1stsolution

x=1

3+5y=15

5y=12

y=12÷5

2ndsolution

x=2

6+5y=15

5y=9

y=9÷5

3rdsolution

y=1

3x+5=15

3x=10

x=10÷3

4thsolution

y=2

3x+10=15

3x=5

x=5÷3

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