the first equation of the system is mutiplied by 4, by what number would you mutiply the second equation to eliminate the y variable by adding? first: 2x+5y=16 second:8x - 4y =10 i got 2 0r 5 im not sure which one
Mulitpy it by 5, so that the coefficent of y in each is 20.
so the answer is negative 5 right
No. It is +5, as Bob correcly stated. The coefficient of y in the second equation then becomes -20, so you can eliminate that variable by adding to the first equation.
To eliminate the y variable by adding, you need the coefficients of y to be equal but with opposite signs in the two equations. In other words, you want the numbers in front of y to be 5 in the first equation and -4 in the second equation.
You correctly multiplied the first equation by 4, which resulted in 4(2x + 5y) = 4(16), giving us 8x + 20y = 64.
Now, you need to determine what number to multiply the second equation by to make the coefficient of y -4. We can achieve this by multiplying the entire equation by -5, since -5 * -4 = 20.
By multiplying the second equation by -5, we get: -5(8x - 4y) = -5(10), which simplifies to -40x + 20y = -50.
Now we have the two equations: 8x + 20y = 64 and -40x + 20y = -50.
Notice that the y terms now have coefficients of 20 and 20, one positive and one negative, which will cancel each other out when combined.
Therefore, the correct number to multiply the second equation by is -5.