Solve the system of equations. {:[3x+4y=-23],[x=3y+1],[x=],[y=]:}

Given Equations: We are given a system of equations: 3x + 4y = -23 x = 3y + 1 We need to find the values of x and y that satisfy both equations simultaneously. Let's use the substitution method to solve the system. We can substitute the expression for x from the second equation into the first equation.Substitute x into first equation: Substitute x = 3y + 1 into the first equation: 3(3y + 1) + 4y = -23 Now, let's expand and simplify the equation. 9y + 3 + 4y = -23 Combine like terms: 13y + 3 = -23Simplify the equation: Subtract 3 from both sides of the equation to isolate the term with y: 13y + 3 - 3 = -23 - 3 13y = -26 Now, divide both sides by 13 to solve for y: y = -26 / 13 y = -2Isolate y term: Now that we have the value of y, we can substitute it back into the second equation to find x: x = 3y + 1 x = 3(-2) + 1 x = -6 + 1 x = -5

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Solve the system of equations.

{:[3x+4y=-23],[x=3y+1],[x=],[y=]:}

Solve the system of equations. $\newline$ $\begin{array}{l} 3 x+4 y=-23 \\ x=3 y+1 \\ x= \\ y= \end{array}$

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Q. Solve the system of equations.

\newline

\begin{array}{l} 3 x+4 y=-23 \\ x=3 y+1 \\ x= \\ y= \end{array}

Given Equations: We are given a system of equations: $\newline$ $3x + 4y = -23$ $\newline$ $x = 3y + 1$ $\newline$ We need to find the values of $x$ and $y$ that satisfy both equations simultaneously. $\newline$ Let's use the substitution method to solve the system. We can substitute the expression for $x$ from the second equation into the first equation.
Substitute $x$ into first equation: Substitute $x = 3y + 1$ into the first equation: $\newline$ $3(3y + 1) + 4y = -23$ $\newline$ Now, let's expand and simplify the equation. $\newline$ $9y + 3 + 4y = -23$ $\newline$ Combine like terms: $\newline$ $13y + 3 = -23$
Simplify the equation: Subtract $3$ from both sides of the equation to isolate the term with $y$ : $\newline$ $13y + 3 - 3 = -23 - 3$ $\newline$ $13y = -26$ $\newline$ Now, divide both sides by $13$ to solve for $y$ : $\newline$ $y = -26 / 13$ $\newline$ $y = -2$
Isolate y term: Now that we have the value of $y$ , we can substitute it back into the second equation to find $x$ :
$x = 3y + 1$
$x = 3(-2) + 1$
$x = -6 + 1$
$x = -5$