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Solve the system by substitution.

{:[-4x+4y=-40],[y=3x-4]:}

Solve the system by substitution.\newline4x+4y=40y=3x4 \begin{aligned} -4 x+4 y & =-40 \\ y & =3 x-4 \end{aligned}

Full solution

Q. Solve the system by substitution.\newline4x+4y=40y=3x4 \begin{aligned} -4 x+4 y & =-40 \\ y & =3 x-4 \end{aligned}
  1. Substitute yy with 3x43x - 4: Substitute yy with 3x43x - 4 in the first equation.\newlineThe first equation is 4x+4y=40-4x + 4y = -40 and the second equation gives us y=3x4y = 3x - 4. We will substitute yy in the first equation with the expression from the second equation.\newline4x+4(3x4)=40-4x + 4(3x - 4) = -40
  2. Distribute and simplify: Distribute 44 into the parentheses and simplify.\newline4x+12x16=40-4x + 12x - 16 = -40\newlineCombine like terms.\newline8x16=408x - 16 = -40
  3. Combine like terms: Add 1616 to both sides of the equation to isolate the term with xx.\newline8x16+16=40+168x - 16 + 16 = -40 + 16\newline8x=248x = -24
  4. Add 1616 to isolate xx: Divide both sides by 88 to solve for xx.\newline8x8=248\frac{8x}{8} = \frac{-24}{8}\newlinex=3x = -3
  5. Divide by 88 for xx: Substitute x=3x = -3 into the second equation to solve for yy. The second equation is y=3x4y = 3x - 4. Substitute 3-3 for xx. y=3(3)4y = 3(-3) - 4
  6. Substitute xx into second equation: Multiply and subtract to find the value of yy.\newliney=94y = -9 - 4\newliney=13y = -13

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