Math  /  Algebra

QuestionSolve the system of linear equations by substitution. y=x72x+3y=14\begin{array}{l} y=x-7 \\ 2 x+3 y=14 \end{array}
If there are an infinite number of solutions, enter IS in each answer field. If there are no solutions, enter NS in each answer x=x= \square y=y= \square

Studdy Solution

STEP 1

1. The system of equations consists of two linear equations: y=x7y = x - 7 and 2x+3y=142x + 3y = 14.
2. We will use the substitution method to solve this system, which involves substituting the expression for yy from the first equation into the second equation.
3. We aim to find numerical values for xx and yy that satisfy both equations.

STEP 2

1. Substitute the expression for yy from the first equation into the second equation.
2. Solve the resulting equation for xx.
3. Substitute the value of xx back into the first equation to solve for yy.
4. Verify the solution by plugging the values of xx and yy back into both original equations.

STEP 3

Substitute y=x7y = x - 7 into the second equation 2x+3y=142x + 3y = 14.
2x+3(x7)=14 2x + 3(x - 7) = 14

STEP 4

Simplify the equation by distributing and combining like terms.
2x+3x21=14 2x + 3x - 21 = 14 5x21=14 5x - 21 = 14

STEP 5

Solve for xx by isolating xx on one side of the equation.
5x21=14 5x - 21 = 14 5x=14+21 5x = 14 + 21 5x=35 5x = 35 x=7 x = 7

STEP 6

Substitute x=7x = 7 back into the first equation y=x7y = x - 7 to find yy.
y=77 y = 7 - 7 y=0 y = 0

STEP 7

Verify the solution by substituting x=7x = 7 and y=0y = 0 back into both original equations.
First equation: y=x7y = x - 7 0=77 0 = 7 - 7 0=0(True) 0 = 0 \quad \text{(True)}
Second equation: 2x+3y=142x + 3y = 14 2(7)+3(0)=14 2(7) + 3(0) = 14 14+0=14 14 + 0 = 14 14=14(True) 14 = 14 \quad \text{(True)}
Since both equations are satisfied, the solution is confirmed.
The solution to the system of equations is:
x=7x = 7 y=0y = 0

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