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What values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] satisfy the system of equations?

[tex]\[
\begin{cases}
3x + 2y = 14 \\
x = 4y - 2
\end{cases}
\][/tex]

A. No solution

B. [tex]\(\left(\frac{50}{11}, \frac{18}{11}\right)\)[/tex]

C. [tex]\(\left(\frac{10}{7}, \frac{25}{7}\right)\)[/tex]

D. Infinitely many solutions

E. [tex]\(\left(\frac{26}{7}, \frac{10}{7}\right)\)[/tex]

F. [tex]\(\left(\frac{18}{11}, \frac{30}{11}\right)\)[/tex]



Answer :

GinnyAnswer
To determine the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy the given system of equations:

[tex]\[ \left\{\begin{array}{l} 3x + 2y = 14 \\ x = 4y - 2 \end{array}\right. \][/tex]

we will proceed step-by-step:

1. Substitute the second equation into the first equation:

The second equation gives [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = 4y - 2 \][/tex]

Substitute [tex]\( x = 4y - 2 \)[/tex] into the first equation:
[tex]\[ 3(4y - 2) + 2y = 14 \][/tex]

2. Simplify and solve for [tex]\( y \)[/tex]:

Expand and combine like terms:
[tex]\[ 12y - 6 + 2y = 14 \][/tex]

Combine the [tex]\( y \)[/tex] terms:
[tex]\[ 14y - 6 = 14 \][/tex]

Add 6 to both sides:
[tex]\[ 14y = 20 \][/tex]

Divide by 14:
[tex]\[ y = \frac{20}{14} = \frac{10}{7} \][/tex]

3. Substitute [tex]\( y = \frac{10}{7} \)[/tex] back into the second equation to solve for [tex]\( x \)[/tex]:

Using [tex]\( x = 4y - 2 \)[/tex]:
[tex]\[ x = 4\left(\frac{10}{7}\right) - 2 \][/tex]
[tex]\[ x = \frac{40}{7} - 2 \][/tex]
[tex]\[ x = \frac{40}{7} - \frac{14}{7} \][/tex]
[tex]\[ x = \frac{26}{7} \][/tex]

Thus, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy the system of equations are:

[tex]\[ \left(\frac{26}{7}, \frac{10}{7}\right) \][/tex]

Therefore, the correct answer is:

[tex]\(\left(\frac{26}{7}, \frac{10}{7}\right)\)[/tex]

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