Question

drag the solution to each system of equations into the appropriate box.
system
$x + \frac{1}{2}y = 3$
$x + y = 5$
$y = 2x + 2$
$y = 4x + 1$
solution
$\left(\frac{1}{2}, 2\
ight)$
$\left(\frac{1}{2}, 3\
ight)$
$\left(\frac{1}{2}, 4\
ight)$
$(1, 3)$
$(, 4)$
$\left(2, \frac{1}{2}\
ight)$
$(2, 1)$
$(2, 2)$
$(1, 2)$

Explanation:

Step1: Solve first system (elimination)

First system:
$x + \frac{1}{2}y = 3$
$x + y = 5$

Subtract first eq from second:
$(x + y) - (x + \frac{1}{2}y) = 5 - 3$
$\frac{1}{2}y = 2$
$y = 4$

Substitute $y=4$ into $x+y=5$:
$x + 4 = 5$
$x = 1$
Solution: $(1, 4)$

Step2: Solve second system (substitution)

Second system:
$y = 2x + 2$
$y = 4x + 1$

Set equal:
$2x + 2 = 4x + 1$
$2 - 1 = 4x - 2x$
$1 = 2x$
$x = \frac{1}{2}$

Substitute $x=\frac{1}{2}$ into $y=2x+2$:
$y = 2(\frac{1}{2}) + 2 = 1 + 2 = 3$
Solution: $(\frac{1}{2}, 3)$

Answer:

System	Solution
$y = 2x + 2$ <br> $y = 4x + 1$	$(\frac{1}{2}, 3)$