Asked by God'sgift Kufre oku
How can I solve the problem 5d+2e=1;8d+3e=2 simultaneously by elimination.
Answers
Answered by
GPT-4o mini
To solve the system of equations:
1. \( 5d + 2e = 1 \) (Equation 1)
2. \( 8d + 3e = 2 \) (Equation 2)
using the elimination method, we will eliminate one of the variables by making the coefficients of one of the variables equal in both equations. Let's eliminate \( e \) in this case.
### Step 1: Make coefficients of \( e \) equal
To do this, we can multiply Equation 1 by \( 3 \) and Equation 2 by \( 2 \) (to make the coefficients of \( e \) equal):
\[
3(5d + 2e) = 3(1) \implies 15d + 6e = 3 \quad \text{(Equation 3)}
\]
\[
2(8d + 3e) = 2(2) \implies 16d + 6e = 4 \quad \text{(Equation 4)}
\]
Now we have:
- Equation 3: \( 15d + 6e = 3 \)
- Equation 4: \( 16d + 6e = 4 \)
### Step 2: Subtract the equations
Next, we subtract Equation 3 from Equation 4 to eliminate \( e \):
\[
(16d + 6e) - (15d + 6e) = 4 - 3
\]
This simplifies to:
\[
d = 1
\]
### Step 3: Substitute back to find \( e \)
Now that we have \( d = 1 \), we can substitute this value back into either Equation 1 or Equation 2 to find \( e \). Let's use Equation 1:
\[
5(1) + 2e = 1
\]
\[
5 + 2e = 1
\]
Subtract 5 from both sides:
\[
2e = 1 - 5
\]
\[
2e = -4
\]
Now divide by 2 to solve for \( e \):
\[
e = -2
\]
### Final Solution
Thus, the solution to the system of equations is:
\[
d = 1, \quad e = -2
\]
You can check the solution by substituting \( d = 1 \) and \( e = -2 \) back into the original equations to ensure both are satisfied.
1. \( 5d + 2e = 1 \) (Equation 1)
2. \( 8d + 3e = 2 \) (Equation 2)
using the elimination method, we will eliminate one of the variables by making the coefficients of one of the variables equal in both equations. Let's eliminate \( e \) in this case.
### Step 1: Make coefficients of \( e \) equal
To do this, we can multiply Equation 1 by \( 3 \) and Equation 2 by \( 2 \) (to make the coefficients of \( e \) equal):
\[
3(5d + 2e) = 3(1) \implies 15d + 6e = 3 \quad \text{(Equation 3)}
\]
\[
2(8d + 3e) = 2(2) \implies 16d + 6e = 4 \quad \text{(Equation 4)}
\]
Now we have:
- Equation 3: \( 15d + 6e = 3 \)
- Equation 4: \( 16d + 6e = 4 \)
### Step 2: Subtract the equations
Next, we subtract Equation 3 from Equation 4 to eliminate \( e \):
\[
(16d + 6e) - (15d + 6e) = 4 - 3
\]
This simplifies to:
\[
d = 1
\]
### Step 3: Substitute back to find \( e \)
Now that we have \( d = 1 \), we can substitute this value back into either Equation 1 or Equation 2 to find \( e \). Let's use Equation 1:
\[
5(1) + 2e = 1
\]
\[
5 + 2e = 1
\]
Subtract 5 from both sides:
\[
2e = 1 - 5
\]
\[
2e = -4
\]
Now divide by 2 to solve for \( e \):
\[
e = -2
\]
### Final Solution
Thus, the solution to the system of equations is:
\[
d = 1, \quad e = -2
\]
You can check the solution by substituting \( d = 1 \) and \( e = -2 \) back into the original equations to ensure both are satisfied.
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