 Hi and welcome to the session. Let's discuss the following question. The question says, from the choices given below, choose the equation whose graphs are given in figure 4.6 and figure 4.7. This is the graph given in figure 4.6 and this is the graph given in figure 4.7. For figure 4.6, first equation is y is equal to x, second is x plus y is equal to 0, third is y is equal to 2x, and fourth is 2 plus 3 y is equal to 7x. And for figure 4.7, first equation is y is equal to x plus 2, second is y is equal to x minus 2, third is y is equal to minus x plus 2, and fourth is x plus 2 y is equal to 6. Before solving this question, we should know that every point of a linear equation variables is a solution in your equation. This is the key idea in this question. Add the solution for figure 4.6. The equation is y is equal to x. We know that every point on the graph of a linear equation in two variables is a solution of the linear equation. A solution means pair of values 1 for x and 1 for y that satisfy the given equation. So in this question, we will verify whether the coordinates which are given on the line satisfy the given equation or not. If this satisfies the given equation, so this means that the graph which is given to us is the graph of that equation, otherwise it is not. The points on the line minus 1, 1, 0, 0, 1 minus 1 substitute minus 1 in the equation y is equal to x. Right. So by substituting x as 1 and y as minus 1, we get left inside as minus 1 and right inside as 1. So this implies that left inside is not equal to right inside. Thus, y is equal to x is not the required equation as x plus y is equal to 0. So let's substitute the points minus 1, 1, 0, 0, 1 minus 1. If I substitute minus 1, 1 then we get LHS as minus 1 plus 1 and this is equal to 0. Our LHS is also equal to 0. Therefore LHS is equal to LHS. Now substitute the point 0, 0. Now by substituting x as 0 and y as 0, we get LHS as 0 plus 0. So our LHS is equal to 0 and our LHS is also equal to 0. Therefore LHS is equal to LHS. Right. And now substitute the point 1 minus 1. So by substituting x as 1 and y as minus 1, we get LHS as 1 minus 1, 1 minus 1 is equal to 0. Our LHS is also equal to 0. Therefore LHS is equal to LHS. Other points satisfy the equation x plus y is equal to 0. So this implies that this graph is of the line x plus y is equal to 0 plus x plus y is equal to 0 is the required equation. This is y is equal to 2x. So again we will substitute the points minus 1, 1, 0, 0 and 1 minus 1 in this equation. So let's first substitute the point minus 1, 1. Now by substituting x as minus 1 and y as 1, we get LHS as 1 and RHS as minus 2. Therefore LHS is not equal to RHS. Right. And thus y is equal to 2x is not required equation. Last equation is 2 plus 3y is equal to 7x. So again we will substitute the point minus 1, 1, 0, 0 and 1 minus 1. So let's first substitute minus 1, 1. Now here x is equal to minus 1 and y is equal to 1. So now substitute these values in this equation. So by substituting the values we get LHS as 2 plus 3 into 1 and this is equal to 5 and our RHS will be equal to 7 into minus 1. So RHS is equal to minus 7. Therefore LHS is not equal to RHS thus plus 3y is equal to 7x is not the required equation. Figure 4.7 first equation is y is equal to x plus 2. So we will again substitute the points minus 1, 3, 0, 2 and 2, 0 in this equation. So let's first substitute minus 1, 3. Now here x is equal to minus 1 and y is equal to 3. So by substituting these values we get LHS as 3 and RHS as minus 1 plus 2 and this implies RHS is equal to 1. Therefore LHS is not equal to RHS. So this implies that the point minus 1, 3 does not satisfy this equation and this implies minus 1, 3 is not a solution of y is equal to x plus 2. So this graph does not represent the equation y is equal to x plus 2. Thus y is equal to x plus 2 is not the required equation. Any equation is y is equal to x minus 2. So let's first substitute minus 1, 3. Now here x is equal to minus 1 and y is equal to 3. So by substituting these values we get LHS as 3 and RHS as minus 1 minus 2 and this is equal to minus 3. Therefore LHS is not equal to RHS and this means that minus 1, 3 is not a solution of y is equal to x minus 2. So this graph does not represent the equation y is equal to x minus 2. Thus y is equal to x minus 2 is not the required equation. Third equation given to us is y is equal to minus x plus 2. So here we will substitute the points minus 1, 3, 0, 2 and 2, 0 in this equation. So let's first substitute minus 1, 3. Now here x is equal to minus 1 and y is equal to 3. So by substituting these values we get LHS as 3 and RHS as therefore LHS is equal to RHS. Now we will substitute the point 0, 2. Now here x is equal to 0 and y is equal to 2. So by substituting these values we get LHS as 2 and RHS as 2. Therefore LHS is equal to RHS. Now substitute the point to 0. Now here x is equal to 2 and y is equal to 0. So by substituting these values we get LHS as 0 and RHS as minus 2 plus 2 is equal to 0. LHS is equal to RHS. Now all the points satisfy this equation. This means minus 1, 3, 0, 2 and 2, 0 is the solution of the equation y is equal to minus x plus 2. So this means that this graph represents the equation y is equal to minus x plus 2. Thus y is equal to minus x plus 2 is the required equation. Last equation is x plus 2 y is equal to 6. So again we will substitute all the points in this equation. So let's first substitute minus 1, 3. Now here x is equal to minus 1 and y is equal to 3. So by substituting these values we get LHS as minus 1 plus 2 into 3 and this is equal to 5 and our RHS is equal to 6. Therefore LHS is not equal to RHS. x plus 2 y is equal to 6 is not the required equation. So for figure 4.6 x plus y is equal to 0 is the required equation and for figure 4.7 y is equal to minus x plus 2 is the required equation. So this completes this question. Bye and take care.