IDNLearn.com provides a seamless experience for finding and sharing answers. Join our Q&A platform to access reliable and detailed answers from experts in various fields.
Sagot :
Certainly! Let's solve each system of equations using the elimination method:
### (a) [tex]\( x + 2y = 8 \)[/tex] and [tex]\( 2x + 3y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} x + 2y = 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
2. Multiply the first equation by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[ \begin{cases} 2(x + 2y) = 2 \cdot 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 2x + 4y = 16 \\ 2x + 3y = 13 \end{cases} \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (2x + 4y) - (2x + 3y) = 16 - 13 \][/tex]
Simplifying, we get:
[tex]\[ y = 3 \][/tex]
4. Substitute [tex]\( y = 3 \)[/tex] back into the first equation:
[tex]\[ x + 2(3) = 8 \][/tex]
Simplifying, we get:
[tex]\[ x + 6 = 8 \implies x = 2 \][/tex]
So, the solution for (a) is [tex]\( x = 2 \)[/tex] and [tex]\( y = 3 \)[/tex].
### (b) [tex]\( 3x - 2y = 7 \)[/tex] and [tex]\( 5x + y = 3 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 5x + y = 3 \end{cases} \][/tex]
2. Multiply the second equation by 2 to align the coefficients of [tex]\( y \)[/tex]:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 2(5x + y) = 2 \cdot 3 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 10x + 2y = 6 \end{cases} \][/tex]
3. Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (3x - 2y) + (10x + 2y) = 7 + 6 \][/tex]
Simplifying, we get:
[tex]\[ 13x = 13 \implies x = 1 \][/tex]
4. Substitute [tex]\( x = 1 \)[/tex] back into the second equation:
[tex]\[ 5(1) + y = 3 \][/tex]
Simplifying, we get:
[tex]\[ 5 + y = 3 \implies y = -2 \][/tex]
So, the solution for (b) is [tex]\( x = 1 \)[/tex] and [tex]\( y = -2 \)[/tex].
### (c) [tex]\( 3x - y = 4 \)[/tex] and [tex]\( x + 2y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - y = 4 \\ x + 2y = 13 \end{cases} \][/tex]
2. Multiply the second equation by 3 to align the coefficients of [tex]\( x \)[/tex]:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3(x + 2y) = 3 \cdot 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3x + 6y = 39 \end{cases} \][/tex]
3. Subtract the first equation from the second to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 6y) - (3x - y) = 39 - 4 \][/tex]
Simplifying, we get:
[tex]\[ 7y = 35 \implies y = 5 \][/tex]
4. Substitute [tex]\( y = 5 \)[/tex] back into the second equation:
[tex]\[ x + 2(5) = 13 \][/tex]
Simplifying, we get:
[tex]\[ x + 10 = 13 \implies x = 3 \][/tex]
So, the solution for (c) is [tex]\( x = 3 \)[/tex] and [tex]\( y = 5 \)[/tex].
### (d) [tex]\( 3x + 4y = 1 \)[/tex] and [tex]\( 3x + 2y = 1 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x + 4y = 1 \\ 3x + 2y = 1 \end{cases} \][/tex]
2. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 4y) - (3x + 2y) = 1 - 1 \][/tex]
Simplifying, we get:
[tex]\[ 2y = 0 \implies y = 0 \][/tex]
3. Substitute [tex]\( y = 0 \)[/tex] back into the first equation:
[tex]\[ 3x + 4(0) = 1 \][/tex]
Simplifying, we get:
[tex]\[ 3x = 1 \implies x = \frac{1}{3} \][/tex]
So, the solution for (d) is [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = 0 \)[/tex].
### (a) [tex]\( x + 2y = 8 \)[/tex] and [tex]\( 2x + 3y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} x + 2y = 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
2. Multiply the first equation by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[ \begin{cases} 2(x + 2y) = 2 \cdot 8 \\ 2x + 3y = 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 2x + 4y = 16 \\ 2x + 3y = 13 \end{cases} \][/tex]
3. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (2x + 4y) - (2x + 3y) = 16 - 13 \][/tex]
Simplifying, we get:
[tex]\[ y = 3 \][/tex]
4. Substitute [tex]\( y = 3 \)[/tex] back into the first equation:
[tex]\[ x + 2(3) = 8 \][/tex]
Simplifying, we get:
[tex]\[ x + 6 = 8 \implies x = 2 \][/tex]
So, the solution for (a) is [tex]\( x = 2 \)[/tex] and [tex]\( y = 3 \)[/tex].
### (b) [tex]\( 3x - 2y = 7 \)[/tex] and [tex]\( 5x + y = 3 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 5x + y = 3 \end{cases} \][/tex]
2. Multiply the second equation by 2 to align the coefficients of [tex]\( y \)[/tex]:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 2(5x + y) = 2 \cdot 3 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - 2y = 7 \\ 10x + 2y = 6 \end{cases} \][/tex]
3. Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (3x - 2y) + (10x + 2y) = 7 + 6 \][/tex]
Simplifying, we get:
[tex]\[ 13x = 13 \implies x = 1 \][/tex]
4. Substitute [tex]\( x = 1 \)[/tex] back into the second equation:
[tex]\[ 5(1) + y = 3 \][/tex]
Simplifying, we get:
[tex]\[ 5 + y = 3 \implies y = -2 \][/tex]
So, the solution for (b) is [tex]\( x = 1 \)[/tex] and [tex]\( y = -2 \)[/tex].
### (c) [tex]\( 3x - y = 4 \)[/tex] and [tex]\( x + 2y = 13 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x - y = 4 \\ x + 2y = 13 \end{cases} \][/tex]
2. Multiply the second equation by 3 to align the coefficients of [tex]\( x \)[/tex]:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3(x + 2y) = 3 \cdot 13 \end{cases} \][/tex]
This simplifies to:
[tex]\[ \begin{cases} 3x - y = 4 \\ 3x + 6y = 39 \end{cases} \][/tex]
3. Subtract the first equation from the second to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 6y) - (3x - y) = 39 - 4 \][/tex]
Simplifying, we get:
[tex]\[ 7y = 35 \implies y = 5 \][/tex]
4. Substitute [tex]\( y = 5 \)[/tex] back into the second equation:
[tex]\[ x + 2(5) = 13 \][/tex]
Simplifying, we get:
[tex]\[ x + 10 = 13 \implies x = 3 \][/tex]
So, the solution for (c) is [tex]\( x = 3 \)[/tex] and [tex]\( y = 5 \)[/tex].
### (d) [tex]\( 3x + 4y = 1 \)[/tex] and [tex]\( 3x + 2y = 1 \)[/tex]
1. Given the equations:
[tex]\[ \begin{cases} 3x + 4y = 1 \\ 3x + 2y = 1 \end{cases} \][/tex]
2. Subtract the second equation from the first to eliminate [tex]\( x \)[/tex]:
[tex]\[ (3x + 4y) - (3x + 2y) = 1 - 1 \][/tex]
Simplifying, we get:
[tex]\[ 2y = 0 \implies y = 0 \][/tex]
3. Substitute [tex]\( y = 0 \)[/tex] back into the first equation:
[tex]\[ 3x + 4(0) = 1 \][/tex]
Simplifying, we get:
[tex]\[ 3x = 1 \implies x = \frac{1}{3} \][/tex]
So, the solution for (d) is [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = 0 \)[/tex].
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
