Answer :
Sure, let's solve these systems of linear equations using the method of elimination.
### Set 1:
(i) [tex]\(2x + 5y = -1\)[/tex]
(ii) [tex]\(x + y = 2\)[/tex]
First, we'll eliminate one of the variables. Let's eliminate [tex]\(x\)[/tex].
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x + y) = 2 \cdot 2 \implies 2x + 2y = 4 \][/tex]
2. Now we have:
[tex]\[ 2x + 5y = -1 \][/tex]
[tex]\[ 2x + 2y = 4 \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 5y) - (2x + 2y) = -1 - 4 \implies 3y = -5 \][/tex]
[tex]\[ y = -\frac{5}{3} \][/tex]
4. Substitute [tex]\(y = -\frac{5}{3}\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x + \left(-\frac{5}{3}\right) = 2 \implies x - \frac{5}{3} = 2 \implies x = 2 + \frac{5}{3} = \frac{6}{3} + \frac{5}{3} = \frac{11}{3} \][/tex]
So, the solution to this system is:
[tex]\[ x = \frac{11}{3}, \quad y = -\frac{5}{3} \][/tex]
### Set 2:
(i) [tex]\(2x + 3y = 3\)[/tex]
(ii) [tex]\(x - 2y = 4\)[/tex]
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x - 2y) = 2 \cdot 4 \implies 2x - 4y = 8 \][/tex]
2. Now we have:
[tex]\[ 2x + 3y = 3 \][/tex]
[tex]\[ 2x - 4y = 8 \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 3y) - (2x - 4y) = 3 - 8 \implies 7y = -5 \][/tex]
[tex]\[ y = -\frac{5}{7} \][/tex]
4. Substitute [tex]\(y = -\frac{5}{7}\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x - 2\left(-\frac{5}{7}\right) = 4 \implies x + \frac{10}{7} = 4 \implies x = 4 - \frac{10}{7} = \frac{28}{7} - \frac{10}{7} = \frac{18}{7} \][/tex]
So, the solution to this system is:
[tex]\[ x = \frac{18}{7}, \quad y = -\frac{5}{7} \][/tex]
### Set 3:
(i) [tex]\(2x + 3y = 3\)[/tex]
(ii) [tex]\(x - y = 0\)[/tex]
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x - y) = 2 \cdot 0 \implies 2x - 2y = 0 \][/tex]
2. Now we have:
[tex]\[ 2x + 3y = 3 \][/tex]
[tex]\[ 2x - 2y = 0 \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 3y) - (2x - 2y) = 3 - 0 \implies 5y = 3 \][/tex]
[tex]\[ y = \frac{3}{5} \][/tex]
4. Substitute [tex]\(y = \frac{3}{5}\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x - \frac{3}{5} = 0 \implies x = \frac{3}{5} \][/tex]
So, the solution to this system is:
[tex]\[ x = \frac{3}{5}, \quad y = \frac{3}{5} \][/tex]
### Set 4:
(i) [tex]\(2x + 3y = 3\)[/tex]
(ii) [tex]\(x + 5y = 5\)[/tex]
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x + 5y) = 2 \cdot 5 \implies 2x + 10y = 10 \][/tex]
2. Now we have:
[tex]\[ 2x + 3y = 3 \][/tex]
[tex]\[ 2x + 10y = 10 \][/tex]
3. Subtract the first equation from the second to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 10y) - (2x + 3y) = 10 - 3 \implies 7y = 7 \][/tex]
[tex]\[ y = 1 \][/tex]
4. Substitute [tex]\(y = 1\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x + 5(1) = 5 \implies x + 5 = 5 \implies x = 0 \][/tex]
So, the solution to this system is:
[tex]\[ x = 0, \quad y = 1 \][/tex]
### Set 1:
(i) [tex]\(2x + 5y = -1\)[/tex]
(ii) [tex]\(x + y = 2\)[/tex]
First, we'll eliminate one of the variables. Let's eliminate [tex]\(x\)[/tex].
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x + y) = 2 \cdot 2 \implies 2x + 2y = 4 \][/tex]
2. Now we have:
[tex]\[ 2x + 5y = -1 \][/tex]
[tex]\[ 2x + 2y = 4 \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 5y) - (2x + 2y) = -1 - 4 \implies 3y = -5 \][/tex]
[tex]\[ y = -\frac{5}{3} \][/tex]
4. Substitute [tex]\(y = -\frac{5}{3}\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x + \left(-\frac{5}{3}\right) = 2 \implies x - \frac{5}{3} = 2 \implies x = 2 + \frac{5}{3} = \frac{6}{3} + \frac{5}{3} = \frac{11}{3} \][/tex]
So, the solution to this system is:
[tex]\[ x = \frac{11}{3}, \quad y = -\frac{5}{3} \][/tex]
### Set 2:
(i) [tex]\(2x + 3y = 3\)[/tex]
(ii) [tex]\(x - 2y = 4\)[/tex]
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x - 2y) = 2 \cdot 4 \implies 2x - 4y = 8 \][/tex]
2. Now we have:
[tex]\[ 2x + 3y = 3 \][/tex]
[tex]\[ 2x - 4y = 8 \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 3y) - (2x - 4y) = 3 - 8 \implies 7y = -5 \][/tex]
[tex]\[ y = -\frac{5}{7} \][/tex]
4. Substitute [tex]\(y = -\frac{5}{7}\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x - 2\left(-\frac{5}{7}\right) = 4 \implies x + \frac{10}{7} = 4 \implies x = 4 - \frac{10}{7} = \frac{28}{7} - \frac{10}{7} = \frac{18}{7} \][/tex]
So, the solution to this system is:
[tex]\[ x = \frac{18}{7}, \quad y = -\frac{5}{7} \][/tex]
### Set 3:
(i) [tex]\(2x + 3y = 3\)[/tex]
(ii) [tex]\(x - y = 0\)[/tex]
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x - y) = 2 \cdot 0 \implies 2x - 2y = 0 \][/tex]
2. Now we have:
[tex]\[ 2x + 3y = 3 \][/tex]
[tex]\[ 2x - 2y = 0 \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 3y) - (2x - 2y) = 3 - 0 \implies 5y = 3 \][/tex]
[tex]\[ y = \frac{3}{5} \][/tex]
4. Substitute [tex]\(y = \frac{3}{5}\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x - \frac{3}{5} = 0 \implies x = \frac{3}{5} \][/tex]
So, the solution to this system is:
[tex]\[ x = \frac{3}{5}, \quad y = \frac{3}{5} \][/tex]
### Set 4:
(i) [tex]\(2x + 3y = 3\)[/tex]
(ii) [tex]\(x + 5y = 5\)[/tex]
1. Multiply equation (ii) by 2 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[ 2(x + 5y) = 2 \cdot 5 \implies 2x + 10y = 10 \][/tex]
2. Now we have:
[tex]\[ 2x + 3y = 3 \][/tex]
[tex]\[ 2x + 10y = 10 \][/tex]
3. Subtract the first equation from the second to eliminate [tex]\(x\)[/tex]:
[tex]\[ (2x + 10y) - (2x + 3y) = 10 - 3 \implies 7y = 7 \][/tex]
[tex]\[ y = 1 \][/tex]
4. Substitute [tex]\(y = 1\)[/tex] back into equation (ii) to find [tex]\(x\)[/tex]:
[tex]\[ x + 5(1) = 5 \implies x + 5 = 5 \implies x = 0 \][/tex]
So, the solution to this system is:
[tex]\[ x = 0, \quad y = 1 \][/tex]
