Solve the following systems of equations: 1) 2x + 3y = 13, 3x + y = 9. 2) 3x − 5y = 7, 2x + y = 9.

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ScanSolve AI Tutor · Accepted Answer

here are the solutions for the two systems of equations.

First System of Equations:

2x + 3y = 13 \quad (1)

3x + y = 9 \quad (2)

Express $y$ in terms of $x$ from equation (2).
From $3x + y = 9$ , we can write:
$y = 9 - 3x$
Substitute this expression for $y$ into equation (1).
$2x + 3(9 - 3x) = 13$ $2x + 27 - 9x = 13$ $-7x + 27 = 13$
Solve for $x$ .
Subtract 27 from both sides:
$-7x = 13 - 27$ $-7x = -14$
Divide by -7:
$x = \frac{-14}{-7}$ $x = 2$
Substitute the value of $x$ back into the expression for $y$ .
$y = 9 - 3(2)$ $y = 9 - 6$ $y = 3$
The solution for the first system is $x = \boxed{2}$ and $y = \boxed{3}$ .

Second System of Equations:
$3x - 5y = 7 \quad (1)$ $2x + y = 9 \quad (2)$
Express $y$ in terms of $x$ from equation (2).
From $2x + y = 9$ , we can write:
$y = 9 - 2x$
Substitute this expression for $y$ into equation (1).
$3x - 5(9 - 2x) = 7$ $3x - 45 + 10x = 7$ $13x - 45 = 7$
Solve for $x$ .
Add 45 to both sides:
$13x = 7 + 45$ $13x = 52$
Divide by 13:
$x = \frac{52}{13}$ $x = 4$
Substitute the value of $x$ back into the expression for $y$ .
$y = 9 - 2(4)$ $y = 9 - 8$ $y = 1$
The solution for the second system is $x = \boxed{4}$ and $y = \boxed{1}$ .

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