2p + 3q = 3100, p + 6q = 1200, 2p + 3y = 23

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for the systems of equations:

System 1:

\begin{align*} 4x + 5y &= 23 \quad &(1) \\ 2x + 3y &= 13 \quad &(2)\end{align*}

Multiply Equation (2) by 2 to make the coefficients of $x$ equal.
$2 \times (2x + 3y) = 2 \times 13$ 4x + 6y = 26 \quad &(3)
Subtract Equation (1) from Equation (3) to eliminate $x$ .
$(4x + 6y) - (4x + 5y) = 26 - 23$ $y = 3$
Substitute the value of $y$ into Equation (2) to find $x$ .
$2x + 3(3) = 13$ $2x + 9 = 13$ $2x = 13 - 9$ $2x = 4$ $x = 2$
The solution for the first system is $\boxed{x=2, y=3}$ .

System 2:
$\begin{align*} p + b &= 1200 \quad &(1) \\ 2p + 3b &= 3100 \quad &(2)\end{align*}$
Multiply Equation (1) by 2 to make the coefficients of $p$ equal.
$2 \times (p + b) = 2 \times 1200$ 2p + 2b = 2400 \quad &(3)
Subtract Equation (3) from Equation (2) to eliminate $p$ .
$(2p + 3b) - (2p + 2b) = 3100 - 2400$ $b = 700$
Substitute the value of $b$ into Equation (1) to find $p$ .
$p + 700 = 1200$ $p = 1200 - 700$ $p = 500$
The solution for the second system is $\boxed{p=500, b=700}$ .

Answer