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Sagot :
To solve the system of linear equations given by:
1. [tex]\( x + 2v = 6 \)[/tex]
2. [tex]\( x - v = 10 \)[/tex]
we need to determine the values of [tex]\( x \)[/tex] and [tex]\( v \)[/tex] that satisfy both equations simultaneously. Here's a step-by-step solution:
### Step 1: Solve one of the equations for one variable.
We'll start by solving the second equation for [tex]\( x \)[/tex].
[tex]\[ x - v = 10 \][/tex]
Add [tex]\( v \)[/tex] to both sides to isolate [tex]\( x \)[/tex]:
[tex]\[ x = 10 + v \][/tex]
### Step 2: Substitute this expression into the other equation.
Now that we have [tex]\( x \)[/tex] expressed in terms of [tex]\( v \)[/tex], we'll substitute [tex]\( x = 10 + v \)[/tex] into the first equation [tex]\( x + 2v = 6 \)[/tex].
[tex]\[ (10 + v) + 2v = 6 \][/tex]
### Step 3: Simplify and solve for [tex]\( v \)[/tex].
Combine like terms:
[tex]\[ 10 + v + 2v = 6 \][/tex]
[tex]\[ 10 + 3v = 6 \][/tex]
Subtract 10 from both sides to isolate the term with [tex]\( v \)[/tex]:
[tex]\[ 3v = 6 - 10 \][/tex]
[tex]\[ 3v = -4 \][/tex]
Now, divide by 3 to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \frac{-4}{3} \][/tex]
### Step 4: Substitute [tex]\( v \)[/tex] back into the expression for [tex]\( x \)[/tex].
We previously derived that [tex]\( x = 10 + v \)[/tex]. Now, substitute [tex]\( v = -\frac{4}{3} \)[/tex] into this equation:
[tex]\[ x = 10 + \left(-\frac{4}{3}\right) \][/tex]
[tex]\[ x = 10 - \frac{4}{3} \][/tex]
To combine these, express 10 as a fraction with a common denominator:
[tex]\[ x = \frac{30}{3} - \frac{4}{3} \][/tex]
[tex]\[ x = \frac{30 - 4}{3} \][/tex]
[tex]\[ x = \frac{26}{3} \][/tex]
### Step 5: Verify the solution.
To ensure the solution is correct, substitute [tex]\( x = \frac{26}{3} \)[/tex] and [tex]\( v = -\frac{4}{3} \)[/tex] back into both original equations to check if they hold true.
First equation:
[tex]\[ x + 2v = 6 \][/tex]
[tex]\[ \frac{26}{3} + 2\left(-\frac{4}{3}\right) = 6 \][/tex]
[tex]\[ \frac{26}{3} - \frac{8}{3} = 6 \][/tex]
[tex]\[ \frac{18}{3} = 6 \][/tex]
[tex]\[ 6 = 6 \][/tex] (This is true.)
Second equation:
[tex]\[ x - v = 10 \][/tex]
[tex]\[ \frac{26}{3} - \left(-\frac{4}{3}\right) = 10 \][/tex]
[tex]\[ \frac{26}{3} + \frac{4}{3} = 10 \][/tex]
[tex]\[ \frac{30}{3} = 10 \][/tex]
[tex]\[ 10 = 10 \][/tex] (This is true.)
Since the solution satisfies both equations, the solution is:
[tex]\[ x = \frac{26}{3}, \quad v = -\frac{4}{3} \][/tex]
1. [tex]\( x + 2v = 6 \)[/tex]
2. [tex]\( x - v = 10 \)[/tex]
we need to determine the values of [tex]\( x \)[/tex] and [tex]\( v \)[/tex] that satisfy both equations simultaneously. Here's a step-by-step solution:
### Step 1: Solve one of the equations for one variable.
We'll start by solving the second equation for [tex]\( x \)[/tex].
[tex]\[ x - v = 10 \][/tex]
Add [tex]\( v \)[/tex] to both sides to isolate [tex]\( x \)[/tex]:
[tex]\[ x = 10 + v \][/tex]
### Step 2: Substitute this expression into the other equation.
Now that we have [tex]\( x \)[/tex] expressed in terms of [tex]\( v \)[/tex], we'll substitute [tex]\( x = 10 + v \)[/tex] into the first equation [tex]\( x + 2v = 6 \)[/tex].
[tex]\[ (10 + v) + 2v = 6 \][/tex]
### Step 3: Simplify and solve for [tex]\( v \)[/tex].
Combine like terms:
[tex]\[ 10 + v + 2v = 6 \][/tex]
[tex]\[ 10 + 3v = 6 \][/tex]
Subtract 10 from both sides to isolate the term with [tex]\( v \)[/tex]:
[tex]\[ 3v = 6 - 10 \][/tex]
[tex]\[ 3v = -4 \][/tex]
Now, divide by 3 to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \frac{-4}{3} \][/tex]
### Step 4: Substitute [tex]\( v \)[/tex] back into the expression for [tex]\( x \)[/tex].
We previously derived that [tex]\( x = 10 + v \)[/tex]. Now, substitute [tex]\( v = -\frac{4}{3} \)[/tex] into this equation:
[tex]\[ x = 10 + \left(-\frac{4}{3}\right) \][/tex]
[tex]\[ x = 10 - \frac{4}{3} \][/tex]
To combine these, express 10 as a fraction with a common denominator:
[tex]\[ x = \frac{30}{3} - \frac{4}{3} \][/tex]
[tex]\[ x = \frac{30 - 4}{3} \][/tex]
[tex]\[ x = \frac{26}{3} \][/tex]
### Step 5: Verify the solution.
To ensure the solution is correct, substitute [tex]\( x = \frac{26}{3} \)[/tex] and [tex]\( v = -\frac{4}{3} \)[/tex] back into both original equations to check if they hold true.
First equation:
[tex]\[ x + 2v = 6 \][/tex]
[tex]\[ \frac{26}{3} + 2\left(-\frac{4}{3}\right) = 6 \][/tex]
[tex]\[ \frac{26}{3} - \frac{8}{3} = 6 \][/tex]
[tex]\[ \frac{18}{3} = 6 \][/tex]
[tex]\[ 6 = 6 \][/tex] (This is true.)
Second equation:
[tex]\[ x - v = 10 \][/tex]
[tex]\[ \frac{26}{3} - \left(-\frac{4}{3}\right) = 10 \][/tex]
[tex]\[ \frac{26}{3} + \frac{4}{3} = 10 \][/tex]
[tex]\[ \frac{30}{3} = 10 \][/tex]
[tex]\[ 10 = 10 \][/tex] (This is true.)
Since the solution satisfies both equations, the solution is:
[tex]\[ x = \frac{26}{3}, \quad v = -\frac{4}{3} \][/tex]
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