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Sagot :
Let's analyze and solve the system of equations step by step.
### First Equation
The first equation given is:
[tex]\[ 49^{2x} \times \left(\frac{1}{7}\right)^{y-4} = \frac{1}{343} \][/tex]
#### Step-by-Step Solution:
1. Rewrite in terms of the same base:
- Notice that [tex]\( 49 = 7^2 \)[/tex] and [tex]\( 343 = 7^3 \)[/tex].
- The equation [tex]\(49^{2x}\)[/tex] can be rewritten as [tex]\((7^2)^{2x} = 7^{4x}\)[/tex].
- The fraction [tex]\(\left(\frac{1}{7}\right)^{y-4}\)[/tex] can be written as [tex]\(7^{-(y-4)}\)[/tex].
- The right-hand side [tex]\(\frac{1}{343}\)[/tex] can be written as [tex]\(7^{-3}\)[/tex].
2. Substitute these values:
[tex]\[ 7^{4x} \times 7^{-(y-4)} = 7^{-3} \][/tex]
3. Combine the exponents:
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we get:
[tex]\[ 7^{4x - (y - 4)} = 7^{-3} \][/tex]
[tex]\[ 7^{4x - y + 4} = 7^{-3} \][/tex]
4. Set the exponents equal to each other:
Since the bases are the same, the exponents must be equal:
[tex]\[ 4x - y + 4 = -3 \][/tex]
5. Simplify the equation:
[tex]\[ 4x - y = -7 \][/tex]
This gives us the first equation in a simplified linear form:
[tex]\[ 4x - y = -7 \][/tex]
Let's call this Equation (1).
### Second Equation
The second equation given is:
[tex]\[ 3^{x+3} \div \frac{1}{279} = \frac{1}{9} \][/tex]
#### Step-by-Step Solution:
1. Rewrite in terms of the same base:
- The equation [tex]\(\frac{1}{279}\)[/tex] can be simplified as [tex]\(3^{-1}\)[/tex] because [tex]\(279 \neq 3^3\)[/tex], let's assume there was a slight mistake here and take the original base from fractional exponents [tex]\( \)[/tex].
- The right-hand side [tex]\(\frac{1}{9}\)[/tex] can be written as [tex]\(3^{-2}\)[/tex].
2. Simplify the equation:
[tex]\[ 3^{x+3} \times 3^{-1} = 3^{-2} \][/tex]
3. Combine the exponents:
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we get:
[tex]\[ 3^{(x+3) + (-1)} = 3^{-2} \][/tex]
[tex]\[ 3^{x + 2} = 3^{-2} \][/tex]
4. Set the exponents equal to each other:
Since the bases are the same, the exponents must be equal:
[tex]\[ x + 2 = -2 \][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = -2 - 2 \][/tex]
[tex]\[ x = -6 \][/tex]
We have the value of [tex]\(x\)[/tex]:
[tex]\[ x = -6 \][/tex]
### Substitute [tex]\(x = -6\)[/tex] into Equation (1)
Using Equation (1):
[tex]\[ 4x - y = -7 \][/tex]
1. Substitute [tex]\(x = -6\)[/tex]:
[tex]\[ 4(-6) - y = -7 \][/tex]
[tex]\[ -24 - y = -7 \][/tex]
2. Solve for [tex]\(y\)[/tex]:
[tex]\[ -y = -7 + 24 \][/tex]
[tex]\[ -y = 17 \][/tex]
[tex]\[ y = -17 \][/tex]
### Final Solution
[tex]\[ x = -6 \][/tex]
[tex]\[ y = -17 \][/tex]
The pair [tex]\((x, y)\)[/tex] that satisfies both equations is [tex]\((-6, -17)\)[/tex].
### First Equation
The first equation given is:
[tex]\[ 49^{2x} \times \left(\frac{1}{7}\right)^{y-4} = \frac{1}{343} \][/tex]
#### Step-by-Step Solution:
1. Rewrite in terms of the same base:
- Notice that [tex]\( 49 = 7^2 \)[/tex] and [tex]\( 343 = 7^3 \)[/tex].
- The equation [tex]\(49^{2x}\)[/tex] can be rewritten as [tex]\((7^2)^{2x} = 7^{4x}\)[/tex].
- The fraction [tex]\(\left(\frac{1}{7}\right)^{y-4}\)[/tex] can be written as [tex]\(7^{-(y-4)}\)[/tex].
- The right-hand side [tex]\(\frac{1}{343}\)[/tex] can be written as [tex]\(7^{-3}\)[/tex].
2. Substitute these values:
[tex]\[ 7^{4x} \times 7^{-(y-4)} = 7^{-3} \][/tex]
3. Combine the exponents:
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we get:
[tex]\[ 7^{4x - (y - 4)} = 7^{-3} \][/tex]
[tex]\[ 7^{4x - y + 4} = 7^{-3} \][/tex]
4. Set the exponents equal to each other:
Since the bases are the same, the exponents must be equal:
[tex]\[ 4x - y + 4 = -3 \][/tex]
5. Simplify the equation:
[tex]\[ 4x - y = -7 \][/tex]
This gives us the first equation in a simplified linear form:
[tex]\[ 4x - y = -7 \][/tex]
Let's call this Equation (1).
### Second Equation
The second equation given is:
[tex]\[ 3^{x+3} \div \frac{1}{279} = \frac{1}{9} \][/tex]
#### Step-by-Step Solution:
1. Rewrite in terms of the same base:
- The equation [tex]\(\frac{1}{279}\)[/tex] can be simplified as [tex]\(3^{-1}\)[/tex] because [tex]\(279 \neq 3^3\)[/tex], let's assume there was a slight mistake here and take the original base from fractional exponents [tex]\( \)[/tex].
- The right-hand side [tex]\(\frac{1}{9}\)[/tex] can be written as [tex]\(3^{-2}\)[/tex].
2. Simplify the equation:
[tex]\[ 3^{x+3} \times 3^{-1} = 3^{-2} \][/tex]
3. Combine the exponents:
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we get:
[tex]\[ 3^{(x+3) + (-1)} = 3^{-2} \][/tex]
[tex]\[ 3^{x + 2} = 3^{-2} \][/tex]
4. Set the exponents equal to each other:
Since the bases are the same, the exponents must be equal:
[tex]\[ x + 2 = -2 \][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = -2 - 2 \][/tex]
[tex]\[ x = -6 \][/tex]
We have the value of [tex]\(x\)[/tex]:
[tex]\[ x = -6 \][/tex]
### Substitute [tex]\(x = -6\)[/tex] into Equation (1)
Using Equation (1):
[tex]\[ 4x - y = -7 \][/tex]
1. Substitute [tex]\(x = -6\)[/tex]:
[tex]\[ 4(-6) - y = -7 \][/tex]
[tex]\[ -24 - y = -7 \][/tex]
2. Solve for [tex]\(y\)[/tex]:
[tex]\[ -y = -7 + 24 \][/tex]
[tex]\[ -y = 17 \][/tex]
[tex]\[ y = -17 \][/tex]
### Final Solution
[tex]\[ x = -6 \][/tex]
[tex]\[ y = -17 \][/tex]
The pair [tex]\((x, y)\)[/tex] that satisfies both equations is [tex]\((-6, -17)\)[/tex].
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