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Sagot :
To solve the given simultaneous linear equations, let's follow these steps:
The equations we are given are:
1. [tex]\(2x + 3y = 1200\)[/tex]
2. [tex]\(5x + 4y = 2300\)[/tex]
Step 1: Identify the equations.
We have two equations:
[tex]\[ 2x + 3y = 1200 \][/tex]
[tex]\[ 5x + 4y = 2300 \][/tex]
Step 2: Use the method of elimination or substitution to solve for one of the variables.
To eliminate one of the variables, we can make the coefficients of [tex]\(y\)[/tex] the same in both equations. Let's multiply the first equation by 4 and the second equation by 3:
[tex]\[ 4(2x + 3y) = 4(1200) \][/tex]
[tex]\[ 8x + 12y = 4800 \][/tex]
[tex]\[ 3(5x + 4y) = 3(2300) \][/tex]
[tex]\[ 15x + 12y = 6900 \][/tex]
Step 3: Subtract one equation from the other to eliminate [tex]\(y\)[/tex].
[tex]\[ (15x + 12y) - (8x + 12y) = 6900 - 4800 \][/tex]
[tex]\[ 15x + 12y - 8x - 12y = 2100 \][/tex]
[tex]\[ 7x = 2100 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex].
[tex]\[ x = \frac{2100}{7} \][/tex]
[tex]\[ x = 300 \][/tex]
Step 5: Substitute [tex]\(x\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex].
Let's substitute [tex]\(x = 300\)[/tex] into the first equation:
[tex]\[ 2(300) + 3y = 1200 \][/tex]
[tex]\[ 600 + 3y = 1200 \][/tex]
[tex]\[ 3y = 1200 - 600 \][/tex]
[tex]\[ 3y = 600 \][/tex]
[tex]\[ y = \frac{600}{3} \][/tex]
[tex]\[ y = 200 \][/tex]
Step 6: State the solution.
The solution to the system of equations is:
[tex]\[ x = 300 \][/tex]
[tex]\[ y = 200 \][/tex]
Thus, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy the given equations are:
[tex]\[ x = 300 \][/tex]
[tex]\[ y = 200 \][/tex]
The equations we are given are:
1. [tex]\(2x + 3y = 1200\)[/tex]
2. [tex]\(5x + 4y = 2300\)[/tex]
Step 1: Identify the equations.
We have two equations:
[tex]\[ 2x + 3y = 1200 \][/tex]
[tex]\[ 5x + 4y = 2300 \][/tex]
Step 2: Use the method of elimination or substitution to solve for one of the variables.
To eliminate one of the variables, we can make the coefficients of [tex]\(y\)[/tex] the same in both equations. Let's multiply the first equation by 4 and the second equation by 3:
[tex]\[ 4(2x + 3y) = 4(1200) \][/tex]
[tex]\[ 8x + 12y = 4800 \][/tex]
[tex]\[ 3(5x + 4y) = 3(2300) \][/tex]
[tex]\[ 15x + 12y = 6900 \][/tex]
Step 3: Subtract one equation from the other to eliminate [tex]\(y\)[/tex].
[tex]\[ (15x + 12y) - (8x + 12y) = 6900 - 4800 \][/tex]
[tex]\[ 15x + 12y - 8x - 12y = 2100 \][/tex]
[tex]\[ 7x = 2100 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex].
[tex]\[ x = \frac{2100}{7} \][/tex]
[tex]\[ x = 300 \][/tex]
Step 5: Substitute [tex]\(x\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex].
Let's substitute [tex]\(x = 300\)[/tex] into the first equation:
[tex]\[ 2(300) + 3y = 1200 \][/tex]
[tex]\[ 600 + 3y = 1200 \][/tex]
[tex]\[ 3y = 1200 - 600 \][/tex]
[tex]\[ 3y = 600 \][/tex]
[tex]\[ y = \frac{600}{3} \][/tex]
[tex]\[ y = 200 \][/tex]
Step 6: State the solution.
The solution to the system of equations is:
[tex]\[ x = 300 \][/tex]
[tex]\[ y = 200 \][/tex]
Thus, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy the given equations are:
[tex]\[ x = 300 \][/tex]
[tex]\[ y = 200 \][/tex]
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