 Hi, and welcome to the session. Let's discuss the following question. The question says, reduce the following equations into intercept form, and find the intercepts or the axels. Third part is 3y plus 2 is equal to 0. The first one in this question, which we know that equation of straight line in intercept form is plus y by v is equal to 1, where a is the x intercept, b is the y intercept. Let us remember that equation of straight line in intercept form is x by a plus y by v is equal to 1, where a is the x intercept, and b is the y intercept. Begin with the solution. Given equation is 3y plus 2 is equal to 0. Now this implies 3y is equal to minus 2 on dividing both sides of this equation by minus 2, we get 3y minus 2y is equal to 1. Now this equation can further be written as x by 0, since x is not present here, so we can write it as x by 0, plus y by 3, because y by 2 by 3 is equal to 3y minus 2y. This equation is of the form x by a plus y by v is equal to 1. So this is the required intercept form of the given equation, 3y plus 2 is equal to 0. Let's name this equation as equation number 1. On comparing equal to 1, we find equal to 0 minus 2 by 3, equal to 0, and b is equal to minus 2 by 3 means y is the given equation is 2 by 3 is equal to 1, I should pi and take care of.