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solve the system of linear equations by substitution x+4y=-1 and -3x-14=y

Sagot :

Explanation

We are given the equations below:

[tex]\begin{gathered} x+4y=-1(equation\text{ }1) \\ -3x-14=y(equation\text{ }2) \end{gathered}[/tex]

We are required to solve the equations above simultaneously using substitution. Thus, we have:

[tex]\begin{gathered} From\text{ }equation\text{ }1, \\ x+4y=-1 \\ Make\text{ }x\text{ }the\text{ }subject \\ x=-1-4y(equation\text{ }3) \\ Substitute\text{ }for\text{ }x\text{ }into\text{ }equation\text{ }2 \\ Equation\text{ }2:-3x-14=y \\ \Rightarrow-3(-1-4y)-14=y \\ 3+12y-14=y \\ 12y-11=y \\ Collect\text{ }like\text{ }terms \\ -11=y-12y \\ -11=-11y \\ \frac{-11}{-11}=\frac{-11y}{-11} \\ y=1 \end{gathered}[/tex][tex]\begin{gathered} From\text{ }equation\text{ }3, \\ x=-1-4y \\ Substitute\text{ }the\text{ }value\text{ }of\text{ }y \\ x=-1-4(1) \\ x=-1-4 \\ x=-5 \end{gathered}[/tex]

Hence, the answer is:

[tex]x=-5;y=1[/tex]