Respuesta :
The system of equations we have is:
[tex]\begin{gathered} \\ \mleft\{\begin{aligned}3x+4y=33 \\ y=x-4\end{aligned}\mright. \end{gathered}[/tex]To solve it, we need to have the variables of both equations on the same side.
Step 1: Substract x from both sides of the second equation
[tex]\begin{gathered} y-x=x-x-4 \\ y-x=-4 \\ -x+y=-4 \end{gathered}[/tex]And now the system is:
[tex]\{\begin{aligned}3x+4y=33 \\ -x+y=-4\end{aligned}[/tex]Step 2: We need to eliminate one variable to solve, for this reason now we multiply the second equation by 3
[tex]\begin{gathered} 3(-x+y=-4) \\ -3x+3y=-12 \end{gathered}[/tex]And we add this with the first equation of the system
[tex]\begin{gathered} 3x+4y=33 \\ -3x+3y=-12 \\ ----------- \\ 0x+7y=21 \end{gathered}[/tex]which is the same as:
[tex]7y=21[/tex]Step 3: solve this previous equation for y, by dividing both sides by 7
[tex]\begin{gathered} \frac{7y}{7}=\frac{21}{7} \\ y=3 \end{gathered}[/tex]Step 4: with the value for y, we find the value of x.
Substituting y=3 in the second equation of the original system of equations:
[tex]y=x-4[/tex][tex]3=x-4[/tex]We add 4 to both sides:
[tex]\begin{gathered} 3+4=x \\ 7=x \end{gathered}[/tex]Answer: y=3 and x=7, representing the result as an ordered pair: (7, 3)
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