Question Solve a system of linear equations with coordinates for the solutions.

Studdy Solution

STEP 1

Assumptions
1. Each pair of equations represents a system of linear equations.
2. Each system of equations must be solved for the variables $x$ and $y$ .
3. The equations are written in the form $(x, y = expression)$ .
4. The numbers outside the parentheses seem to be incorrectly placed and should be ignored for solving the systems.
5. The equations within the parentheses are the ones that need to be solved.

STEP 2

Let's begin by solving the system represented by point A.
The equation for point A is given as: $-0.5y + 20 - 0.5y = 13$

STEP 3

Combine like terms in the equation for point A.
$-0.5y - 0.5y + 20 = 13$

STEP 4

Simplify the equation for point A by combining the $y$ terms.
$-1y + 20 = 13$

STEP 5

Subtract 20 from both sides of the equation for point A to isolate the $y$ term.
$-1y = 13 - 20$

STEP 6

Calculate the value of $y$ for point A.
$-1y = -7$

STEP 7

Divide by -1 to solve for $y$ in the equation for point A.
$y = \frac{-7}{-1}$

STEP 8

Find the value of $y$ for point A.
$y = 7$

STEP 9

Since point A only has an equation for $y$ , we cannot find a corresponding $x$ value. Thus, we only have a value for $y$ for point A.
$A(6, y) = A(6, 7)$

STEP 10

Now, let's solve the system represented by point B.
The equation for point B is given as: $4 - 3x - 7x = -8$

STEP 11

Combine like terms in the equation for point B.
$4 - 10x = -8$

STEP 12

Subtract 4 from both sides of the equation for point B to isolate the $x$ term.
$-10x = -8 - 4$

STEP 13

Calculate the value of $x$ for point B.
$-10x = -12$

STEP 14

Divide by -10 to solve for $x$ in the equation for point B.
$x = \frac{-12}{-10}$

STEP 15

Simplify the fraction to find the value of $x$ for point B.
$x = \frac{6}{5}$

STEP 16

Since point B only has an equation for $x$ , we cannot find a corresponding $y$ value. Thus, we only have a value for $x$ for point B.
$B(x, -8) = B(\frac{6}{5}, -8)$

STEP 17

Now, let's solve the system represented by point H.
The equation for point H is given as: $4y + 14 - 2y = 4$

STEP 18

Combine like terms in the equation for point H.
$4y - 2y + 14 = 4$

STEP 19

Simplify the equation for point H by combining the $y$ terms.
$2y + 14 = 4$

STEP 20

Subtract 14 from both sides of the equation for point H to isolate the $y$ term.
$2y = 4 - 14$

STEP 21

Calculate the value of $y$ for point H.
$2y = -10$

STEP 22

Divide by 2 to solve for $y$ in the equation for point H.
$y = \frac{-10}{2}$

STEP 23

Find the value of $y$ for point H.
$y = -5$

STEP 24

Since point H only has an equation for $y$ , we cannot find a corresponding $x$ value. Thus, we only have a value for $y$ for point H.
$H(-5, y) = H(-5, -5)$

STEP 25

Now, let's solve the system represented by point M.
The equation for point M is given as: $-5y + 10 - y = -2$

STEP 26

Combine like terms in the equation for point M.
$-5y - y + 10 = -2$

STEP 27

Simplify the equation for point M by combining the $y$ terms.
$-6y + 10 = -2$

STEP 28

Subtract 10 from both sides of the equation for point M to isolate the $y$ term.
$-6y = -2 - 10$

STEP 29

Calculate the value of $y$ for point M.
$-6y = -12$

STEP 30

Divide by -6 to solve for $y$ in the equation for point M.
$y = \frac{-12}{-6}$

STEP 31

Find the value of $y$ for point M.
$y = 2$

STEP 32

Since point M only has an equation for $y$ , we cannot find a corresponding $x$ value. Thus, we only have a value for $y$ for point M.
$M(-3, y) = M(-3, 2)$
The solutions for the points are: $A(6, 7), B(\frac{6}{5}, -8), H(-5, -5), M(-3, 2)$

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