Question

you can use substitution to solve a system of equations.
to use substitution, one of the equations needs to have a variable alone on one side. in this system, neither equation has a variable alone.
$x - 3y = 16$
$x + 2y = 26$
solve either equation for one of the variables. choose the equation you think would be easier to solve. lets solve the first equation for $x$. add $3y$ to both sides.
$x - 3y = 16$
$x = 16 + 3y$
since $x = 16 + 3y$, you can substitute $16 + 3y$ for $x$ in the second equation. then solve for $y$.
$x + 2y = 26$
$16 + 3y + 2y = 26$
$16 + 5y = 26$
$5y = 10$
$y = 2$
now that you know $y$, you can find $x$. substitute $2$ for $y$ in either equation to solve for $x$. lets use the second equation, $x + 2y = 26$.
$x + 2y = 26$
$x + 2(2) = 26$
$x + 4 = 26$
$x = 22$
finally, write the solution as an ordered pair. since $x = 22$ and $y = 2$, the solution is $(22, 2)$.
practice! solve each system of equations using substitution.
$y = 3x$
$x + y = 20$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$x = 3$
$-5x + 2y = 1$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$3x + 5y = 4$
$y = -x - 2$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$y = 6x - 12$
$y = -6x$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$x - 2y = 22$
$x + y = 10$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$y = 3x + 12$
$y = -4x + 5$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$2x - y = 1$
$3x + 4y = 40$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$x + 2y = -16$
$3x - 5y = -15$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
$5x + 7y = 16$
$2x + y = 10$
$(\\_\\_\\_\\_, \\_\\_\\_\\_)$
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Explanation:

Step1: Solve \( y = 3x \) and \( x + y = 20 \)

Substitute \( y = 3x \) into \( x + y = 20 \):
\( x + 3x = 20 \)
\( 4x = 20 \)
\( x = 5 \)
Then \( y = 3(5) = 15 \).

Step2: Solve \( x = 3 \) and \( -5x + 2y = 1 \)

Substitute \( x = 3 \) into \( -5x + 2y = 1 \):
\( -5(3) + 2y = 1 \)
\( -15 + 2y = 1 \)
\( 2y = 16 \)
\( y = 8 \).

Step3: Solve \( 3x + 5y = 4 \) and \( y = -x - 2 \)

Substitute \( y = -x - 2 \) into \( 3x + 5y = 4 \):
\( 3x + 5(-x - 2) = 4 \)
\( 3x - 5x - 10 = 4 \)
\( -2x = 14 \)
\( x = -7 \)
Then \( y = -(-7) - 2 = 5 \).

Step4: Solve \( y = 6x - 12 \) and \( y = -6x \)

Set \( 6x - 12 = -6x \):
\( 12x = 12 \)
\( x = 1 \)
Then \( y = -6(1) = -6 \).

Step5: Solve \( x - 2y = 22 \) and \( x + y = 10 \)

From \( x + y = 10 \), \( x = 10 - y \). Substitute into \( x - 2y = 22 \):
\( 10 - y - 2y = 22 \)
\( 10 - 3y = 22 \)
\( -3y = 12 \)
\( y = -4 \)
Then \( x = 10 - (-4) = 14 \).

Step6: Solve \( y = 3x + 12 \) and \( y = -4x + 5 \)

Set \( 3x + 12 = -4x + 5 \):
\( 7x = -7 \)
\( x = -1 \)
Then \( y = 3(-1) + 12 = 9 \).

Step7: Solve \( 2x - y = 1 \) and \( 3x + 4y = 40 \)

From \( 2x - y = 1 \), \( y = 2x - 1 \). Substitute into \( 3x + 4y = 40 \):
\( 3x + 4(2x - 1) = 40 \)
\( 3x + 8x - 4 = 40 \)
\( 11x = 44 \)
\( x = 4 \)
Then \( y = 2(4) - 1 = 7 \).

Step8: Solve \( x + 2y = -16 \) and \( 3x - 5y = -15 \)

From \( x + 2y = -16 \), \( x = -16 - 2y \). Substitute into \( 3x - 5y = -15 \):
\( 3(-16 - 2y) - 5y = -15 \)
\( -48 - 6y - 5y = -15 \)
\( -11y = 33 \)
\( y = -3 \)
Then \( x = -16 - 2(-3) = -10 \).

Step9: Solve \( 5x + 7y = 16 \) and \( 2x + y = 10 \)

From \( 2x + y = 10 \), \( y = 10 - 2x \). Substitute into \( 5x + 7y = 16 \):
\( 5x + 7(10 - 2x) = 16 \)
\( 5x + 70 - 14x = 16 \)
\( -9x = -54 \)
\( x = 6 \)
Then \( y = 10 - 2(6) = -2 \).

Answer:

(5, 15)
(3, 8)
(-7, 5)
(1, -6)
(14, -4)
(-1, 9)
(4, 7)
(-10, -3)
(6, -2)