Question

Find the two solutions for: (4x+7y=28)​

Answer 1

Step-by-step explanation:

Answer :

Answer :• Point of intercept on x-axis is A(7 , 0) .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,The given equation is : 4x + 7y = 28 .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,The given equation is : 4x + 7y = 28 .• At x-axis , y-coordinate is zero .

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,The given equation is : 4x + 7y = 28 .• At x-axis , y-coordinate is zero .Thus ,Putting y = 0 in the given equation ;

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,The given equation is : 4x + 7y = 28 .• At x-axis , y-coordinate is zero .Thus ,Putting y = 0 in the given equation ;We have ;

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,The given equation is : 4x + 7y = 28 .• At x-axis , y-coordinate is zero .Thus ,Putting y = 0 in the given equation ;We have ;=> 4x + 7•0 = 28

Answer :• Point of intercept on x-axis is A(7 , 0) .• Point of intercept on y-axis is B(0 , 4) .Working rule :→ Choose two or three points satisfying the given equation .→ Mark those points on the graph paper .→ Join those points with scale .Solution :Here ,The given equation is : 4x + 7y = 28 .• At x-axis , y-coordinate is zero .Thus ,Putting y = 0 in the given equation ;We have ;=> 4x + 7•0 = 28=> 4x + 0 = 28

=> 4x = 28

=> 4x = 28=> x = 28/4

=> x = 7

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .• At y-axis , x-coordinate is zero .

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .• At y-axis , x-coordinate is zero .Thus ,Putting x = 0 in the given equation

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .• At y-axis , x-coordinate is zero .Thus ,Putting x = 0 in the given equation We have ;

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .• At y-axis , x-coordinate is zero .Thus ,Putting x = 0 in the given equation We have ;=> 4•0 + 7y = 28

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .• At y-axis , x-coordinate is zero .Thus ,Putting x = 0 in the given equation We have ;=> 4•0 + 7y = 28=> 0 + 7y = 28

=> x = 7Hence ,The graph of the given equation will cut the x-axis at the point A(7 , 0) .• At y-axis , x-coordinate is zero .Thus ,Putting x = 0 in the given equation We have ;=> 4•0 + 7y = 28=> 0 + 7y = 28=> 7y = 28

=> y = 28/7

=> y = 4

=> y = 4Hence ,

=> y = 4Hence ,The graph of the given equation will cut the y-axis at the point B(0 , 4) .