Question

Several students attempted to solve the system of equations and . Which one is correct? An image of a coordinate plane shows two lines. A line labeled negative 2x plus y equals 4 crosses the x-axis at (negative 2, 0) and the y-axis at (0, 4). A line labeled x plus y equals 1 crosses the y-axis at (0, 1) and the x-axis at (1, 0). The lines intersect at a point that is one unit left of the origin and two units up from the origin. The solution is . An image of a coordinate plane shows two lines. A line labeled negative 2x plus y equals 4 crosses the y-axis at (0, 4) and the x-axis at (2, 0). A line labeled x plus y equals 1 crosses the x-axis at (negative 1, 0) and the x-axis at (1, 1). The lines intersect at a point that is 1 unit right of the origin and 2 units up from the origin. The solution is . An image of a coordinate plane shows two lines. A line labeled negative 2x plus y equals 4 crosses the y-axis at (0, 4) and the x-axis at (2, 0). A line labeled x plus y equals 1 crosses the x-axis at (negative 1, 0) and the y-axis at (0, 1). The lines intersect at a point that is 1 unit right of the origin and 2 units up from the origin. The solution is . An image of a coordinate plane shows two lines. A line labeled negative 2x plus y equals 4 crosses the x-axis at (negative 2, 0) and the y-axis at (0, 4). A line labeled x plus y equals 1 crosses the y-axis at (0, 1) and the x-axis at (1, 0). The lines intersect at a point that is one unit left of the origin and two units up from the origin. The solution is .

Answer 1

the correct answer would be slope standard form and congruent triangular formula

Answer 2

The student that is correct is the student with the solution (-1, 2), which corresponds to the solution presented in the first option,

The steps used to find the correct solution can be presented as follows;

The system of equations can be presented as follows;

-2·x + y = 4

x + y = 1

The above system of equations can be evaluated as follows;

-2·x + y = 4...(1)

x + y = 1...(2)

Making y the subject of the equation (2), we get;

x + y = 1

y = 1 - x

Plugging in the above value of y into equation (1), we get;

-2·x + y = 4

-2·x + (1 - x) = 4

-3·x + 1 = 4

-3·x = 4 - 1

-3·x = 3

x = 3/(-3)

x = -1

y = 1 - (-1)

y = 2

The solution to the system equation is (-1, 2), which is the first option