Kalikylnnking8408idn
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Solve each system by substitution.
y=5x-7
-3x-2y=-12
Type in your solution (x,y)

Sagot :

[tex] \huge \color{pink} \boxed{AnSwEr \: ♧}[/tex]

Step-by-step explanation:

[tex]\\ \\ \large \color{blue}{Given \: Question↣} \pink{Solve \: by \: substitution \: method} \\ \\ \pink{two \: equations \: given↣y = 5x - 7; - 3x - 2y = - 12} \\ \\ \large \color{blue} {lets \: solve : - } \\ \\ \large \sf \green{we \: know \: that \: y = 5x - 7} \\ \\ \large \sf \green{putting \: y = 5x - 7 \: in \: equation( - 3x - 2y = - 12)} \\ \\ \large \red{ ⇒} \large \sf \green{ \: - 3x - 2(5x - 7) = - 12} \\ \\ \large \red{ ⇒} \large \sf \green{- 3x - 10x + 14 = - 12} \\ \\\large \red{ ⇒} \large \sf \green{arranging \: terms} \\ \\ \large \red{ ⇒} \large \sf \green{ - 3x - 10x = - 12 - 14} \\ \\ \large \red{ ⇒} \large \sf \green{ - 13 x = - 26} \\ \\ \large \red{ ⇒} \large \sf \green{x = \cancel \frac{ - 26}{ - 13} } \\ \\ \large \red{ ⇒} \large \sf \green{x = 2} \green {(soo \: x = 2)}\\ \\ \large \sf\pink{now \: putting \: x = 2 \: in \: equation \: y = 5x - 7} \\ \\ \\ \\ \large \red{ ⇒} \large \sf \green{y = 5(2) - 7}\\ \\ \large \red{ ⇒} \large \sf \green{y = 10 - 7}\\ \\ \large \red{ ⇒} \large \sf \green{y=3} \\ \\ \\ \huge \color{pink}\frak {hence \: x = 2 \: nd \: y = 3}[/tex]

[tex]\\\\[/tex]

[tex]\large\color{cyan}{\boxed{\mathfrak{❥\:Glossy\:Pearl}}}[/tex]

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