a Let's first solve (1). Since the

x -

terms have opposite coefficients, we will use the Elimination Method to solve the system of equations.

{y = - x + 8 y = x - 2 (I) (II)

SysEqnAdd

$(II):$ Add $(I)$

{y = - x + 8 y + y = x - 2 + (- x + 8)

▼

(II):

Solve for

y

RemovePar

$(II):$ Remove parentheses

{y = - x + 8 y + y = x - 2 - x + 8

AddSubTerms

$(II):$ Add and subtract terms

{y = - x + 8 2 y = 6

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

{y = - x + 8 y = 3

Having solved for

y,

we can find

x

by substituting this value for

y

in the first equation.

{y = - x + 8 y = 3

Substitute

$(I):$ $y = 3$

{3 = - x + 8 y = 3

▼

(I):

Solve for

x

AddEqn

$(I):$ $LHS + x = RHS + x$

{x + 3 = 8 y = 3

SubEqn

$(I):$ $LHS - 3 = RHS - 3$

{x = 5 y = 3

The solution to (1) is

(5, 3) .

Let's also solve (2). Again, since both equations have

y

on the left-hand side and with opposite coefficients, we should use the Elimination Method.

{2 x - y = 10 y = - 4 x + 2 (I) (II)

SysEqnAdd

$(II):$ Add $(I)$

{2 x - y = 10 y + (2 x - y) = - 4 x + 2 + 10

▼

(II):

Solve for

x

RemovePar

$(II):$ Remove parentheses

{2 x - y = 10 y + 2 x - y = - 4 x + 2 + 10

AddSubTerms

$(II):$ Add and subtract terms

{2 x - y = 10 2 x = - 4 x + 12

AddEqn

$(II):$ $LHS + 4 x = RHS + 4 x$

{2 x - y = 10 6 x = 12

DivEqn

$(II):$ $LHS / 6 = RHS / 6$

{y = - x + 8 x = 2

We have found that

x = 2

. By substituting this value for

x

in the second equation, we can find

y .

{2 x - y = 10 x = 2

Substitute

$(I):$ $x = 2$

{2 (2) - y = 10 x = 2

▼

(I):

Solve for

y

Multiply

$(I):$ Multiply

{4 - y = 10 x = 2

SubEqn

$(I):$ $LHS - 4 = RHS - 4$

{- y = 6 x = 2

MultEqn

$(I):$ $LHS \cdot (- 1) = RHS \cdot (- 1)$

{y = - 6 x = 2

The solution to (2) is

(2, - 6) .

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