The system of equations has the solution (4, -2). To test our solution, we will substitute this point into the original system and simplify. If the left-hand side and right-hand side are equal, then we have the correct solution.

3x-y=14 & (I) x=2y+8 & (II)

(I), (II): x= 4, y= -2

3( 4)-( -2)? =14 & (I) 4? =2( -2)+8 & (II)

(I), (II): Multiply

12-(-2)? =14 4? =-4+8

SubNeg

(I): a-(- b)=a+b

12+2? =14 4? =-4+8

(I), (II): Add terms

14=14 ✓ 4 = 4 ✓

The solution is correct.

b Since the first equation is already solved for x, we should use the Substitution Method to solve the system of equations.

x=2y+2 & (I) x=- y -10 & (II)

Substitute

(II): x= 2y+2

x=2y+2 2y+2=- y -10

▼

(II): Solve for y

AddEqn

(II): LHS+y=RHS+y

x=2y+2 3y+2=-10

SubEqn

(II): LHS-2=RHS-2

x=2y+2 3y=-12

DivEqn

(II): .LHS /3.=.RHS /3.

x=2y+2 y=-4

Having solved for y, we can substitute this into the second equation and solve for x.

x=2y+2 y=-4

Substitute

(I): y= - 4

x=2( -4)+2 y=-4

MultPosNeg

(I): a(- b)=- a * b

x=-8+2 y=-4

AddTerms

(I): Add terms

x=-6 y=-4

The system of equations has the solution (-6, -4). To test our solution, we will substitute this point into the original system and simplify. If the left-hand side and right-hand side are equal, then we have the correct solution.

x=2y+2 & (I) x=- y -10 & (II)

(I), (II): x= -6, y= -4

-6? =2( -4)+2 & (I) -6? =- ( -4) -10 & (II)

NegNeg

(II): - (- a)=a

-6? =2(-4)+2 -6? =4-10

MultPosNeg

(I): a(- b)=- a * b

-6? =-8+2 -6? =4-10

(I), (II): Add and subtract terms

-6=-6 ✓ -6=-6 ✓

The solution is correct.

c Since the coefficients of y in both equations have opposite signs, we can solve the system using the Elimination Method — we can add the equations to each other to eliminate the variable y.

16x-y=-4 & (I) 2x+y=13 & (II)

SysEqnAdd

(II): Add (I)

16x-y=-4 & (I) 2x+y+ 16x-y=13+( -4) & (II)

▼

(II): Solve for x

AddNeg

(II): a+(- b)=a-b

16x-y=-4 & (I) 2x+y+16x-y=13-4 & (II)

AddSubTerms

(II): Add and subtract terms

16x-y=-4 & (I) 18x=9 & (II)

DivEqn

(II): .LHS /18.=.RHS /18.

16x-y=-4 & (I) x= 12 & (II)

Having solved for x, we can substitute this into the second equation and solve for y.

16x-y=-4 & (I) x= 12 & (II)

Substitute

(I): x= 1/2

16( 12)-y=-4 & (I) x= 12 & (II)

▼

(I): Solve for y

MoveLeftFacToNumOne

(I): a* 1/b= a/b

162-y=-4 & (I) x= 12 & (II)

CalcQuot

(I): Calculate quotient

8-y=-4 & (I) x= 12 & (II)

AddEqn

(I): LHS+4=RHS+4

12-y=0 & (I) x= 12 & (II)

AddEqn

(I): LHS+y=RHS+y

12=y & (I) x= 12 & (II)

RearrangeEqn

(I): Rearrange equation

y=12 & (I) x= 12 & (II)

The system of equations has the solution ( 12, 12). To test our solution, we will substitute this point into the original system and simplify. If the left-hand side and right-hand side are equal, then we have the correct solution.

16x-y=-4 & (I) 2x+y=13 & (II)

(I), (II): x= 1/2, y= 12

16( 12)- 12? =-4 & (I) 2( 12)+ 12? =13 & (II)

(I), (II): a* 1/b= a/b

162- 12? =-4 & (I) 22+ 12? =13 & (II)

(I), (II): Calculate quotient

8- 12? =-4 & (I) 1+ 12? =13 & (II)

(I), (II): Add and subtract terms

-4=-4 ✓ 13=13 ✓

The solution is correct.

Subchapter links

9.2.1 Quadratic Applications with Inequalities p.502-503

9.2.2 Solving Systems of Equations p.507-508

Exercise 59 Page 503

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