solve it---Question Number: 6 Chart Description: Type: Cartesian coordinate system graph. Main Elements: - X-axis is labeled 'x' and extends horizontally. It is marked with integers from -1 to 9. - Y-axis is labeled 'y' and extends vertically. It is marked with integers from -2 to 5. - The origin is labeled 'O' at (0, 0). - A grid is present, with lines corresponding to the integer markings on the axes. - Two thick black lines are drawn on the graph, representing two linear equations. - The two lines intersect at a single point within the first quadrant. - The intersection point appears to be at coordinates (4, 1). Textual Information: The graph of a system of linear equations is shown. The solution to the system is (x, y). What is the value of x ?

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In this problem, we're given a system of linear equations represented by two lines on a coordinate plane. We need to find the value of x in the solution. The solution to a system of linear equations is the point where the lines intersect. Looking at the graph, we can see that the two lines intersect at the point with coordinates 4 comma 1. Therefore, the value of x is 4. Let's find the equations of the two lines in our system. For the first line shown in blue, we can identify two points: the y-intercept at (0, -1) and the intersection point at (4, 1). Using these points, we can calculate the slope as rise over run, which gives us 2 divided by 4, or 1/2. So the equation of this line is y equals one-half x minus 1. For the second line shown in green, we have the points (0, 3) and (4, 1). The slope is negative 2 divided by 4, or negative 1/2. Therefore, the equation is y equals negative one-half x plus 3. These two equations form our system of linear equations. Now, let's solve the system of equations algebraically to verify our graphical solution. We have the two equations: y equals one-half x minus 1, and y equals negative one-half x plus 3. Since both equations equal y, we can set them equal to each other. This gives us one-half x minus 1 equals negative one-half x plus 3. Moving all terms with x to the left side, we get one-half x plus one-half x equals 3 plus 1. Simplifying, we have x equals 4. To find y, we substitute x equals 4 into either equation. Using the first equation, y equals one-half times 4 minus 1, which gives us 2 minus 1, equals 1. Therefore, the solution is the point (4, 1), which confirms our graphical solution. The value of x is 4. Let's solve this system using the substitution method, which is another algebraic approach. We have equation 1: y equals one-half x minus 1, and equation 2: y equals negative one-half x plus 3. In step 1, we substitute the expression for y from equation 1 into equation 2. This gives us one-half x minus 1 equals negative one-half x plus 3. In step 2, we solve for x by moving all terms with x to the left side and all constants to the right side. We get one-half x plus one-half x equals 3 plus 1, which simplifies to x equals 4. In step 3, we substitute this value back into equation 1 to find y. We get y equals one-half times 4 minus 1, which equals 2 minus 1, equals 1. Therefore, the solution is the point (4, 1), and the answer to our question is x equals 4. Let's summarize what we've learned about solving systems of linear equations. A system of linear equations can be solved graphically by finding the intersection point of the lines on a coordinate plane. We can also solve it algebraically using methods like substitution or elimination. For the given system with the equations y equals one-half x minus 1 and y equals negative one-half x plus 3, we found that the solution is the point with coordinates 4 comma 1. Therefore, the answer to the question 'What is the value of x?' is 4. This demonstrates how we can use both graphical and algebraic approaches to solve systems of linear equations.