Solve 5 medals 3x - 5y = 17 9x - 15y = -4
Multiply the top by -3 to solve by elimination. You get -9x + 15y = -51 9x + 15y = -4 \[\Huge 0 \ne -51\] no solution
how do you know when you have to add both x and y termns?
Oops the second equation is 9x - 15 y = -4 * Well you have common terms in both equations.. 3x and 9x both are factors of 3. for example.
you can cancel x out by multiplying the top out by multiplying the equation by -3. but this also cancels out the y term
Something like 3x + 5y = 0 -3x - 5y = 0 0 = 0 this would mean infinite solutions in this case 0 is not equal to -51, so there is no solution
5x + 2y = 7 10x + 4y = 14 --------------- 15x + 6y = 21 -15 15 --------------- 6y=6?
did i do this one right?
No, you have two variables. you cant just get rid of x by -15..
so there's no solution?
Divide the second equation by -2 you get 5x + 2y = 7 -5x -2y = -7 add the equations 0 = 0 this system has infinite solutions
What is the value of y in the solution to the following system of equations? 5x - 3y = -3 2x - 6y = -6
so here what would i do?
Get rid of x. Multiply the top equation by 2 and the bottom by -5 2(5x-3y = -3) -5(2x-6y = -6) 10x - 6y = -6 -10x + 30y = 30 add 24y = 24 y = 1
Use the substitution method to solve the following system of equations. 3x + 5y = 3 x + 2y = 0
so here i would add 3x+5y=3 x+2y=0 ---------- 4x+7y=3
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