Exercise 19 Page 207

Since neither equation has a variable with a coefficient of 1, the Substitution Method may not be the easiest. Instead, we will use the Elimination Method. To do this, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. 3 y-5 x=0 & (I) 2 y-4 x=-2 & (II) Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (I) by -2 and multiply (II) by 3, the y-terms will have opposite coefficients. -2(3 y-5 x)=-2(0) 3(2 y-4 x)=3(-2) ⇒ -6y+10 x=0 6y-12 x=-6 We can see that the y-terms will eliminate each other if we add (I) to (II).

Now we can solve for y by substituting the value of x into Equation (I) and simplifying.

The solution, or intersection point, of the system of equations is (3,5).