can anyone solve this question by substitution method plz ?

Question

4x-y²=-1
x-y=5

Answer 1

x = 5 - y

4(5 - y) - y^2 = -1
20 - 4y - y^2 = -1
21 - 4y - y^2 = 0 or y^2 + 4y - 21 = 0
(y + 7)(y - 3) = 0
y = -7, 3
Use both values of y to solve for x
x -(-7) = 5
x + 7 = 5
x = -2

x - 3 = 5
x = 8

Answer 2

4x - y² = -1 (1)
x-y = 5 (2)

Multiply (2) by -4 and add (1) and (2)

4x - y² = -1
-4x +4y = -20

4y - y² = -21
y²-4y-21 = 0

Sum of the roots = -b/a = -(-4)/1 = 4
Product of the roots = ac = 1 (-21) = -21
where a , b ,c are coefficients of the quadratic equation in the form ax² + bx + c = 0

Roots = (-b ± (√b² - 4ac))/2a where a = 1 b = -4 and c = - 21

(-(-4) ± (√(-4)² - 4(1)(-21))/2(1)
= (4 ± (√16+84))/2
= (4 ± (√100))/2
= (4 ± 10)/2
= (4 + 10)/2 or (4 - 10)/2
= 14/2 or -6/2
= 7 or -3
therefore y = 7 or -3
Sum of the roots = 7-3 = 4
Product of the roots = 7 * -3 = -21

If y = 7 x = y+5 = 12 (2)
if y = - 3 x = y+5 = 2 (2)


Solution (x,y) = (12,7) or (2, -3)

Answer 3

4x-y²=-1
x-y=5 | x=5+y | y=x-5

First, substitute the value of x into the equation, and complete.

4x-y²=-1
4(5+y)-y²=-1
20+4y-y²+1=0
-y²+4y+21=0
-1(-y²+4y+21=0)
y²-4y-21=0
(y-7)(y+3)=0
y-7=0 | y+3=0
y=7 | y=-3

Next, using the x=5+y, you can work out the values of x

x=5+y
x=5+7 | x=5-3
x=12 | x=2

So, the solutions are, (7,12) and (-3,2)

I hope that helps.

Answer 4

4x - y² = -1 .......... (eqn1)
x - y = 5 ....... (eqn 2)
From eqn (2)
x = y + 5
Substituting this value of x in eqn (1) we get

4 ( y + 5 ) - y^2 = - 1
On simplification of this eqn, we get -

y^2 - 4 y - 21 = 0
ie y^2 - 7y + 3y - 21 = 0
ie y ( y - 7 ) + 3 ( y - 7 ) + 0
ie (y - 7) (y + 3) = 0
Hence we get -

y = -3 or y = 7. Now substituting these values we get -
When y = - 3 , x = 2
and for y = 7 , x = 12

Soln Set x : [ 2 , 12 ] , y : [ - 3 , 7 ] ------ Answer

Answer 5

4x-y^2+1=0 (1)

x-y=5
<=>x=5+y (2)

(1) & (2) ==>4x-y^2+1=0
<=>4(5+y) -y^2 +1=0
<=>y=-3 or y= 7
<=>x=2 or x=12

We have two sets of solution (x,y)=(2,-3) or (12,7)

Answer 6

5+y=x 4(5+y)-y2=-1 20+4y-y2=-1 +1+20+4y-y2=0 21+4y-y2=0 -y=5-x -1(y)=-1(5-x) y=-5+x substitute

Answer 7

ok so if you wanna solve it in sub method here's the solution:

1-Solve for x in the second problem so it would be x=5+y
2-Sub (5+y) for 'x' in the first equation... s it would be 4(5+y)-y^2=-1
3-Solve it for 'y'.
4-Sub whatever you have for 'y' in the first equation into any equation for 'y' and then solve for 'x'

Answer 8

x = 5 + y
4(y + 5) - y² = - 1
4y + 20 - y² = - 1
y² - 4y - 21 = 0
(y - 7) (y + 3) = 0
y = 7 , y = - 3
x = 12 , x = 2

(12,7) , (2,-3)

Answer 9

i got 100 for my answer. 5X5=25x4=100.

Answer 10

i will ans it tommorow. just wait.