Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
