3x-2y=-5
3y-4x=8
1) multiply the first equation by 3
3(3x-2y=-5) this will give you the new equation: 9x-6y=-15
2) multiply the second equation by 2
2(3y-4x=8) this should give you the equation: 6y-8x=16
3) combine both equations/ like terms
9x-6y=-15
6y-8x=16
4) -6y and 6y cancel out
9x=-15
-8x=16
5) 9x and -8x combine to make 1x or just x and -15 combined with 16 gives you just 1
6) we are now left with:
x=1
7) plug in the x to any of the two original equations ( i chose the first)
3x-2y=-5
3(1) - 2y = -5
3 - 2y = -5
-2y = -8
y = 4
When you plug in the x=1 you are given 3(1) - 2y = -5
Distribute the 3 and you should have 3 - 2y = -5
Subtract 3 from 3 (this cancels out) then from -5
This should leave you with -2y = -8 ( -3 and -5 add to -8)
Divide by -2 ( -2 divided by -2 cancels out)
-8 divided by -2 gives you 4 (two negatives make a positive)
So, y=4 and x=1
To check, plug in x=1 and y=4 into one equation. when you're done with that you can plug them into the other. when you plug them into the first equation you get -5=-5 which means they worked. when plugged into the second, the result is 8=8 which means x=1 and y=4 worked for both equations.
hello : <span>
<span>the discriminat of each quadratic equation :
ax²+bx+c=0 ....(a ≠ 0) is :
Δ = b² -4ac
1 ) Δ > 0 the equation has two reals solutions : x1,2 =
(-b±√Δ)/2a</span></span>
factoring :
<span>ax²+bx+c = a(x-x1)(x-x2)
2 ) Δ = 0 : one solution : x1 = x2 = -b/2a</span>
factoring :
<span>ax²+bx+c = a(x-x1)²
3 ) Δ < 0 : no reals solutions... no </span>
<span>factoring </span>
<span><span>3/5(50)
=(3/5)(50/1)
=(3)(50)/(5)(1)
=150/5
=30 </span></span>
Answer:
add the = problome
Step-by-step explanation:
Answer:
air because there is no question
