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What is the solution of the system defined by [tex]\( y = -x + 5 \)[/tex] and [tex]\( 5x + 2y = 14 \)[/tex]?

A. [tex]\( x = \frac{1}{3}, y = \frac{11}{7} \)[/tex]
B. [tex]\( x = \frac{4}{3}, y = \frac{11}{3} \)[/tex]
C. [tex]\( x = \frac{5}{11}, y = 0 \)[/tex]
D. [tex]\( x = \frac{4}{3}, y = 1 \)[/tex]


Sagot :

To solve the system of equations
[tex]\[ y = -x + 5 \][/tex]
and
[tex]\[ 5x + 2y = 14, \][/tex]
we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously.

### Step 1: Substitute [tex]\( y \)[/tex] from the first equation into the second equation
The first equation gives us:
[tex]\[ y = -x + 5. \][/tex]

We substitute this expression for [tex]\( y \)[/tex] into the second equation:
[tex]\[ 5x + 2(-x + 5) = 14. \][/tex]

### Step 2: Simplify and solve for [tex]\( x \)[/tex]
Distribute the [tex]\( 2 \)[/tex] in the second equation:
[tex]\[ 5x + 2(-x) + 2(5) = 14. \][/tex]
[tex]\[ 5x - 2x + 10 = 14. \][/tex]

Combine like terms:
[tex]\[ 3x + 10 = 14. \][/tex]

Subtract 10 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 3x = 4. \][/tex]

Divide by 3 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{4}{3}. \][/tex]

### Step 3: Substitute [tex]\( x \)[/tex] back into the first equation to solve for [tex]\( y \)[/tex]
Now we know [tex]\( x = \frac{4}{3} \)[/tex]. Substitute this value back into the first equation:
[tex]\[ y = -\left(\frac{4}{3}\right) + 5. \][/tex]

Convert 5 to a fraction with a common denominator:
[tex]\[ y = -\left(\frac{4}{3}\right) + \left(\frac{15}{3}\right). \][/tex]

Simplify:
[tex]\[ y = \frac{15}{3} - \frac{4}{3} = \frac{11}{3}. \][/tex]

### Conclusion
The solution to the system of equations is:
[tex]\[ x = \frac{4}{3}, \][/tex]
[tex]\[ y = \frac{11}{3}. \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{B. \left( x = \frac{4}{3}, y = \frac{11}{3} \right) }. \][/tex]
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