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Solve the system of equations.

[tex]\[
\begin{array}{l}
7y + 10x = -11 \\
4y - 3x = -15 \\
x = \square \\
y = \square
\end{array}
\][/tex]


Sagot :

To solve the system of equations:
[tex]\[ \begin{array}{l} 7 y + 10 x = -11 \\ 4 y - 3 x = -15 \\ \end{array} \][/tex]

we will use either substitution or elimination. Here, I will detail the elimination method.

### Step 1: Align the equations
First, we write down the two given equations as they are:
[tex]\[ 7 y + 10 x = -11 \tag{1} \][/tex]
[tex]\[ 4 y - 3 x = -15 \tag{2} \][/tex]

### Step 2: Eliminate one variable
To eliminate one of the variables, we can create equivalent coefficients for either [tex]\(x\)[/tex] or [tex]\(y\)[/tex]. In this case, let's eliminate [tex]\(y\)[/tex].

Multiply equation (1) by 4 and equation (2) by 7:
[tex]\[ 4 \cdot (7 y + 10 x) = 4 \cdot (-11) \][/tex]
[tex]\[ 28 y + 40 x = -44 \tag{3} \][/tex]

[tex]\[ 7 \cdot (4 y - 3 x) = 7 \cdot (-15) \][/tex]
[tex]\[ 28 y - 21 x = -105 \tag{4} \][/tex]

### Step 3: Subtract the equations
Next, subtract equation (4) from equation (3):
[tex]\[ (28 y + 40 x) - (28 y - 21 x) = -44 - (-105) \][/tex]

This simplifies to:
[tex]\[ 28 y + 40 x - 28 y + 21 x = -44 + 105 \][/tex]
[tex]\[ 61 x = 61 \][/tex]

### Step 4: Solve for [tex]\(x\)[/tex]
Divide both sides by 61:
[tex]\[ x = 1 \][/tex]

### Step 5: Substitute back to find [tex]\(y\)[/tex]
Now substitute [tex]\(x = 1\)[/tex] back into one of the original equations. Let's use equation (1):
[tex]\[ 7 y + 10 (1) = -11 \][/tex]
[tex]\[ 7 y + 10 = -11 \][/tex]
Subtract 10 from both sides:
[tex]\[ 7 y = -21 \][/tex]
Divide both sides by 7:
[tex]\[ y = -3 \][/tex]

### Final Answer
Thus, the solution to the system of equations is:
[tex]\[ x = 1 \][/tex]
[tex]\[ y = -3 \][/tex]
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