Solve the system of equations. {:[15 x+31 y=-3],[x=-y+3],[x=◻],[y=◻]:}

Given System of Equations: We are given a system of equations: \[ \begin{cases} 15x + 31y = -3 \\ x = -y + 3 \end{cases} \] We will use the second equation to express x in terms of y and then substitute it into the first equation.Express x in terms of y: From the second equation, we have: \[ x = -y + 3 \] Now we will substitute this expression for x into the first equation.Substitute x into first equation: Substituting $ x = -y + 3 $ into the first equation, we get: \[ 15(-y + 3) + 31y = -3 \] Now we will distribute the 15 and simplify the equation.Distribute and simplify: Distributing the 15, we get: \[ -15y + 45 + 31y = -3 \] Now we combine like terms.Combine like terms: Combining like terms, we get: \[ 16y + 45 = -3 \] Next, we will subtract 45 from both sides to isolate the term with y.Isolate y term: Subtracting 45 from both sides, we have: \[ 16y = -3 - 45 \] \[ 16y = -48 \] Now we will divide both sides by 16 to solve for y.Solve for y: Dividing both sides by 16, we get: \[ y = \frac{-48}{16} \] \[ y = -3 \] We have found the value of y. Now we will substitute this value back into the second equation to find x.Substitute y into second equation: Substituting $ y = -3 $ into the second equation $ x = -y + 3 $, we get: \[ x = -(-3) + 3 \] \[ x = 3 + 3 \] \[ x = 6 \] We have found the value of x.

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Solve the system of equations.

{:[15 x+31 y=-3],[x=-y+3],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} 15 x+31 y=-3 \\ x=-y+3 \\ x=\square \\ y=\square \end{array}$

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Algebra 1

Solve a system of equations using substitution

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Q. Solve the system of equations.

\newline

\begin{array}{l} 15 x+31 y=-3 \\ x=-y+3 \\ x=\square \\ y=\square \end{array}

Given System of Equations: We are given a system of equations: $\newline$ $\begin{cases} 15x + 31y = -3 \\ x = -y + 3 \end{cases}$ $\newline$ We will use the second equation to express x in terms of y and then substitute it into the first equation.
Express x in terms of y: From the second equation, we have: $\newline$ $x = -y + 3$ $\newline$ Now we will substitute this expression for x into the first equation.
Substitute x into first equation: Substituting $x = -y + 3$ into the first equation, we get: $\newline$ $15(-y + 3) + 31y = -3$ $\newline$ Now we will distribute the $15$ and simplify the equation.
Distribute and simplify: Distributing the $15$ , we get: $\newline$ $-15y + 45 + 31y = -3$ $\newline$ Now we combine like terms.
Combine like terms: Combining like terms, we get: $\newline$ $16y + 45 = -3$ $\newline$ Next, we will subtract $45$ from both sides to isolate the term with y.
Isolate y term: Subtracting $45$ from both sides, we have: $\newline$ $16y = -3 - 45$ $\newline$ $16y = -48$ $\newline$ Now we will divide both sides by $16$ to solve for y.
Solve for y: Dividing both sides by $16$ , we get: $\newline$ $y = \frac{-48}{16}$ $\newline$ $y = -3$ $\newline$ We have found the value of y. Now we will substitute this value back into the second equation to find x.
Substitute y into second equation: Substituting $y = -3$ into the second equation $x = -y + 3$ , we get: $\newline$ $x = -(-3) + 3$ $\newline$ $x = 3 + 3$ $\newline$ $x = 6$ $\newline$ We have found the value of x.